Luogu quiz: cyclic structure

Today's task, five questions a day, is the result of four days, and I plan to write it in Python and C.
Question source: https://www.luogu.com.cn/training/102#problems

---

1. [Deep basis 4. Example 2] Find the minimum value

Question link: https://www.luogu.com.cn/problem/P5718

Question description

gives $n$ and$n$ integers$a_i$, please this $What\; is\; the\; smallest\; value\; among\; n$ integers?

Input format

The first line enters a positive integer $n$ , represents the number of numbers.

Enter $n$ non-negative integers, representing$a_1,a_2\dots a_n$, separated by spaces.

Output format

Output a non-negative integer representing this nnThe smallest value among n non-negative integers$.$

Example #1

Sample input #1

8
1 9 2 6 0 8 1 7

Sample output #1

0

hint

Data guaranteed, $n\le 100one0$ ≤ ai ≤ 1000 0\ and$0\le a_i\le 1000$。

code(python)

num = int(input())

a = list(map(int, input().split()))
a.sort()
print(a[0])

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>

int main()
{
    
    
	int n;
	scanf("%d",&n);
	
	int min = 1000;
	for(int i = 1;i <= n;i++)
	{
    
    
		int num;
		scanf("%d",&num);
		if(num < min)
		{
    
    
			min = num;
		}
	}
	
	printf("%d\n",min);
	return 0;	
} 

2. [Deep Base 4. Example 3] Classification Average

Question link: https://www.luogu.com.cn/problem/P5719

Question description

Given $n$ and$k$ , will go from 1 tonnAll positive integers between$n$$k$ is divisible (that is,$multiples\; of\; k$ ), while class B numbers cannot. Please output the average of these two types of numbers, accurate to$1$ digit, separated by spaces.

The data guarantees that the number of both types of numbers will not be $0$。

Input format

Enter two positive integers $n$ and$k$。

Output format

Output one line, two real numbers, representing the average number of class A and class B respectively. Accurate to one decimal place.

Example #1

Sample input #1

100 16

Sample output #1

56.0 50.1

hint

Data guarantee, $1\le n\le 10000$，$1\le k\le 100$。

code(python)

# num = int(input())
#
# a = list(map(int, input().split()))
# a.sort()
# print(a[0])

n, k = map(int, input().split(" "))
i = 1
B_num = 0
B_sum = 0
A_num = 0
A_sum = 0
while i <= n:
    if i % k == 0:
        A_num += 1
        A_sum += i
        i += 1
    else:
        B_sum += i
        B_num += 1
        i += 1

print(round(A_sum/A_num, 1),round(B_sum/B_num, 1))

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>

int main()
{
    
    
	int n, k;
	scanf("%d%d",&n,&k);
	
	int x = 0,y = 0;
	int sum = 0,num = 0;
	for(int i = 1;i <= n;i++)
	{
    
    
		if(i % k == 0)
		{
    
    
			sum = sum + i;
			x++;
		}
		if(i % k != 0)
		{
    
    
			num = num + i;
			y++;
		}
	}
	double a = 1.0*sum/x*1.0;
	double b = 1.0*num/y*1.0;
	printf("%.1lf %.1lf",a,b);
	return 0;	
} 

3. [Deep Foundation 4.Example 4] One-foot Chai

Question link: https://www.luogu.com.cn/problem/P5720

Question description

"Zhuangzi" said, "If you take half of a one-foot stick every day, it will be inexhaustible for eternity." On the first day there is a length of aaStarting from the next day, half of the stick must be cut off every day (dividing2 2$each\; time$$2$ , rounded down). On the next few days, the length of the stick will become$1$？

Input format

Enter a positive integer $a$ , represents the length of the stick.

Output format

Output a positive integer, indicating the number of days when the length of the stick will become $1$。

Example #1

Sample input #1

100

Sample output #1

7

hint

Data guarantee, $1\le a\le 10^9$。

code(python)

# num = int(input())
#
# a = list(map(int, input().split()))
# a.sort()
# print(a[0])
import math

n = int(input())
sum = 1
half = n
while half != 1:

    half = int(half / 2)
    sum += 1
print(sum)

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>
 
int main()
{
    
    
	long long n;
	scanf("%lld",&n);
	
	int sum = 1;
	long long half = n;
	while(half !=1)
	{
    
    
			half = half/2;
			sum++;

		
	}
	
	printf("%d\n",sum);
	return 0;	
} 

4. [Deep basis 4. Example 6] Digital right triangle

Question link: https://www.luogu.com.cn/problem/P5721

Question description

gives $n$ , please output a right angle whose side length is$n$ number right triangle. All numbers are2 2Composed of$2$$2$ digits add leading$0$。

Input format

Enter a positive integer $n$。

Output format

Output the numerical right triangle as required by the question.

Example #1

Sample input #1

5

Sample output #1

0102030405
06070809
101112
1314
15

hint

Data guarantee, $1\le n\le 13$。

code(python)

# num = int(input())
#
# a = list(map(int, input().split()))
# a.sort()
# print(a[0])
import math

n = int(input())
i = n
x = 1
while i >= 0:
    j = i - 1
    # print("i:",i)
    while j >= 0:
        # print("j:", j)
        if x < 10:
            print("0%d" % x, end="")
            x += 1
        else:
            print("%d" % x, end="")
            x += 1
        j -= 1
    print("")
    i -= 1

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>
 
int main()
{
    
    
	int n;
	scanf("%d",&n);
	int x = 1;
	for(int i = n;i >= 0;i--)
	{
    
    
		for(int j = i-1;j >= 0;j--)
		{
    
    
			
			if(x < 10)
			{
    
    
				printf("0%d",x);
				x++;
			}
			else
			{
    
    
				printf("%d",x);
				x++;
			}
		}
		printf("\n");
	}
	return 0;	
} 

5. [NOIP1998 Popularization Group] Sum of factorials

Question link: https://www.luogu.com.cn/problem/P1009

Question description

Calculate with high accuracy $S=1!+2!+3!+\cdots +n!$（$n\le 50$）。

where !represents factorial, defined as $n!=n\times (n-1)\times (n-2)\times \cdots \times 1$ . For example,$5!=5\times 4\times 3\times 2\times 1=120$。

Input format

a positive integer $n$。

Output format

a positive integer $S$ , represents the calculation result.

Example #1

Sample input #1

3

Sample output #1

9

hint

【data range】

For $100\%$ of the data,$1\le n\le 50$。

【other instructions】

Note: This question is used as an example in "In-depth Basics", but its data range is only $n\le 20$ , this question cannot be passed using the code in the book.

If you want to pass this question, please continue to study the high-precision knowledge in Chapter 8.

code(python)

n = int(input())
b = 0
for i in range(1, n + 1):
    c = i
    a = 1
    while c != 0:
        a = a * c
        c = c - 1
    b = b + a
print(b)

Code (C)

#include<stdio.h>
int main() {
    
    
	int n;
	int temp;
	int i, j;//进行阶乘和求和时的临时变量
	while(scanf("%d", &n)!=EOF){
    
    
		int m[100] = {
    
    0};//存储阶乘结果
		int num[1000] = {
    
    0};//存储求和结果
		int len = 1, count = 0, C = 0, D = 0;//len为阶乘结果的位数，count为求和结果的位数；C为求阶乘时的进位，D为求和时的进位
		m[0] = 1;//0的阶乘为1
		for (i = 1; i <= n; i++) {
    
    
			for (j = 0; j < len; j++) {
    
    
				temp = m[j];
				m[j] = (temp * i + C) % 10;
				C = (temp * i + C) / 10;
			}
			while (C != 0) {
    
    
				m[len] = C % 10;
				C = C / 10;
				len++;
			}//以上两个循环用来求阶乘结果
			for (count = 0; count < len; count++) {
    
    //求完阶乘直接加
				temp = num[count];
				num[count] = (temp + m[count] + D) % 10;
				D = (temp + m[count] + D) / 10;
			}
			while (D != 0) {
    
    
				m[count + 1] += D % 10;
				D = D / 10;
				count++;
			}//以上两个循环用来求阶乘之和结果
		}
		for (i = count - 1; i >= 0; i--) 
			printf("%d", num[i]);
			printf("\n");
	}
	return 0;
}

6. [NOIP2013 Popularization Group] Counting Issues

Question link: https://www.luogu.com.cn/problem/P1980

Question description

Try to calculate in the interval $1$ tonnAmong all integers of$n$$x$（$0\le x\le 9$ ) How many times does it appear in total? For example, in$1$ to$In\; 11$ , that is, in$1,2,3,4,5,6,7,8,9,10,In\; 11$ , the number$1$ appeared$4$ times.

Input format

$2$ integers$n,x$ , separated by a space.

Output format

$1$ integer, representing$The\; number\; of\; times\; x$ appears.

Example #1

Sample input #1

11 1

Sample output #1

4

hint

For $100\%$ of the data,$1\le n\le 10^6$，$0\le x\le 9$。

code(python)

n,x = list(map(int,input().split()))
# 计数的量
c = 0
# 整数序列
for i in range(1,n+1):
    # count() 计数
    c += str(i).count(str(x))
print(c)

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>

int main(){
    
     
	int n, m;
	scanf("%d%d",&n,&m);
	int num, sum;
	for(int i = 1;i <= n;i++)
	{
    
    
		int pot = i;
		while(pot != 0)
		{
    
    
			num = pot%10;
			pot = pot/10;
			if(num == m)	sum++;
		}
	}
	
	printf("%d",sum);
	return 0;
}

7. [NOIP2002 Popularization Group] Series Summation

Question link: https://www.luogu.com.cn/problem/P1035

Question description

Known: $S_n=1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{n}$. Obviously for any integer $k$ , when$When\; n$ is large enough,$S_n>k$。

Now given an integer $k$ , required to calculate a minimum$n$ , such that$S_n>k$。

Input format

a positive integer $k$。

Output format

a positive integer $n$。

Example #1

Sample input #1

1

Sample output #1

2

hint

【data range】

For $100\%$ of the data,$1\le k\le 15$。

[Source of the question]

NOIP 2002 Popularization Group First Question

code(python)

Sn = 0
k = int(input())
i = 1
while True:
    Sn = Sn+1.0/i
    if Sn > k:
        print(i)
        break
    i += 1

Code (C)

#include <stdio.h>
int main(){
    
    
    double Sn=0;
    int k;
    scanf("%d",&k);
    int i=1;
    while (1)
    {
    
    
        Sn = Sn+1.0/i;
        if(Sn>k)
        {
    
    
            printf("%d",i);
            break;
        }
        i++;
    }

8. [NOIP2015 Popularization Group] Gold Coins

Question link: https://www.luogu.com.cn/problem/P2669

Question background

NOIP2015 popularization group T1

Question description

The king paid gold coins as wages to loyal knights. On the first day, the knight received one gold coin; for the next two days (the second and third days), he received two gold coins every day; for the next three days (the fourth, fifth, and sixth days), he received three gold coins every day. ;In the next four days (the seventh, eighth, ninth, and tenth day), four gold coins will be received every day...; This wage payment model will continue like this: when nn consecutiveReceivednn$every\; day\; for\; n\; days$After$n$$n+In\; 1$ day,$n+1$ gold coin.

Please count before $In\; k$ days, how many gold coins did the knight get in total?

Input format

a positive integer $k$ , represents the number of days to issue gold coins.

Output format

A positive integer, the number of gold coins the knight received.

Example #1

Sample input #1

6

Sample output #1

14

Example #2

Sample input #2

1000

Sample output #2

29820

hint

[Explanation of Sample 1]

The knight receives one gold coin on the first day; two gold coins on the second and third days; three gold coins on the fourth, fifth and sixth days. Therefore, a total of $1+2+2+3+3+3=14$ gold coins.

For $100\%$ of the data,$1\le k\le 10^4$。

code(python)

k = int(input())
n,m = 1,0
while k > 0:
    k -= n
    if k > 0:
        m += n*n
        n += 1
    else:
        m += n*(k+n)
print(m)

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>
int sqr[101];
int app[20001];

int main()
{
    
    
    int k;//k天
	scanf("%d",&k);
	int num = 0;
	int n = 0;
	while(k > num)
	{
    
    	
		n++;
		num = (n+1)*n/2;
		
	} 
	long long sum = 0;
	if(num != k){
    
    
		for(int i = 1;i < n;i++)
		{
    
    
			sum = sum + i*i;
		}
		int m = (n-1)*n/2;
		sum = (k - m)*n + sum;
		printf("%lld\n",sum);
	}
	else if(num == k)
	{
    
    
		for(int i = 1;i <= n;i++)
		{
    
    
			sum = sum + i*i;
		}
		printf("%d\n",sum);
	}
	
	
    return 0;
}

9. [Deep basis 4. Example 11] Sum of a sequence

Question link: https://www.luogu.com.cn/problem/P5722

Question description

Calculate $1+2+3+\cdots +(n-1)+The\; value\; of\; n$ , where the positive integer$n$ is not greater than 100. Since you are not as smart as Gauss, you are not allowed to use the arithmetic sequence summation formula to find the answer directly.

Input format

Enter a positive integer $n$。

Output format

Output a positive integer representing the final summed answer.

Example #1

Sample input #1

100

Sample output #1

5050

hint

Data guarantee, $1\le n\le 100$。

code(python)

n = int(input())
sum1 = 0
for i in range(1, n+1):
    sum1 = sum1 + i

print(sum1)

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>
 
int main()
{
    
    
	int n;
	scanf("%d",&n);
	
	int sum = 0;
	for(int i = 1;i <= n;i++)
	{
    
    
		sum = sum + i;
	}
	
	printf("%d",sum);
	return 0;	
} 

10. [Deep basis 4. Example 13] Prime number pocket

Question link: https://www.luogu.com.cn/problem/P5723

Question description

Little A has a prime number pocket, which can contain various prime numbers. He went from $Starting\; from\; 2$ , it determines whether each natural number is a prime number in turn. If it is a prime number, it will put the number in the pocket.

The load capacity of a pocket is the sum of all the numbers in the pocket.

However, the load-bearing capacity of the pocket is limited, and the sum of the prime numbers cannot exceed $L.$ _ gives$L$ , how many prime numbers can you fit in your pocket? Output these prime numbers from small to large, and then output the maximum number of prime numbers that can be accommodated. Use newlines to separate the numbers.

Input format

One positive integer $L$。

Output format

Output these prime numbers from small to large, and then output the maximum number of prime numbers that can be accommodated, with a blank line between all numbers.

Example #1

Sample input #1

100

Sample output #1

2
3
5
7
11
13
17
19
23
9

Example #2

Sample input #2

5

Sample output #2

2
3
2

Example #3

Sample input #3

11

Sample output #3

2
3
5
3

hint

Data guarantee, $1\le L\le {10}^5$。

code(python)

lena = 0
cnt = 0
prime = [1 for i in range(100010)]
lena = int(input())
prime[1] = 0
prime[0] = 0
for i in range(2, 100010):

    if prime[i] == 1:

        j = i * 2
        while j < 100010:
            prime[j] = 0
            j += i

for i in range(2, 100010):

    if prime[i] == 1:
        if lena >= i:

            print(i)

            lena -= i

            cnt += 1
        else:
            break

print(cnt)

Code (C)

#include <stdio.h>
#include <stdlib.h>

int n,x;
long long sum=0;
int pd(int y) {
    
    
	int i;
	for(i=2; i*i<=y; ++i) {
    
    
		if(y%i==0) return 0;
	}
	return 1;
}
int main() {
    
    
	scanf("%d",&n);
	if(n<2) {
    
    
		printf("0\n");
		return 0;
	} else if(n==2) {
    
    
		printf("2\n1\n");
		return 0;
	}
	int i;
	for(i=2; i<=n; ++i) {
    
    
		if(i%2==0&&i!=2) continue;
		if(sum+i>n) {
    
    
			printf("%d\n",x);
			return 0;
		}
		if(pd(i)) {
    
    
			printf("%d\n",i);
			sum+=i;
			x++;
		}
	}
	return 0;
}

11. [USACO1.5] Prime Palindromes

Question link: https://www.luogu.com.cn/problem/P1217

Question description

Because $151$ is both a prime number and a palindrome (looking the same from left to right and right to left), so$151$ is a palindromic prime number.

Write a program to find the range $[a,b](5\le a<b\le 100,000,000)$ (one hundred million) all palindromic prime numbers.

Input format

The first line enters two positive integers $a$ and$b$。

Output format

Output a list of palindromic primes, one per line.

Example #1

Sample input #1

5 500

Sample output #1

5
7
11
101
131
151
181
191
313
353
373
383

hint

Hint 1: Generate the palindromes and see if they are prime.

Tip 1: Find all the palindromes and determine whether they are prime numbers (prime numbers).

Hint 2: Generate palindromes by combining digits properly. You might need more than one of the loops like below.

Tip 2: To generate the correct number of palindromes, you may need several loops like the one below.

The title translation comes from NOCOW.

USACO Training Section 1.5

produces a length of $The\; number\; of\; palindromes\; of\; 5$ :

for (d1 = 1; d1 <= 9; d1+=2) {
    
        // 只有奇数才会是素数
     for (d2 = 0; d2 <= 9; d2++) {
    
    
         for (d3 = 0; d3 <= 9; d3++) {
    
    
           palindrome = 10000*d1 + 1000*d2 +100*d3 + 10*d2 + d1;//(处理回文数...)
         }
     }
 }

code(python)

# 输入范围
begin, end = map(int, input().split())


# 定义一个筛选素数的函数,时间复杂度为根号n/3
def prime_number(number):
    if number <= 3:
        return True
    elif number % 2 == 0 or number % 3 == 0:
        return False
    else:
        i = 5
        while i * i <= number:
            if number % i == 0 or number % (i + 2) == 0:
                return False
            i += 6
    return True


# 定义生成回文数的函数
from itertools import product


def palind(n):
    digit_palind = []
    half = product(*([range(1, 10, 2)] + [range(10)] * ((n - 1) // 2)))
    for i in half:
        digit_palind.append(n * '%s' % tuple(list(i) + list(i[-(n % 2) - 1::-1])))
    return digit_palind


# 判断终点的数位
numerical_digit = 1
while 10 ** numerical_digit < end:
    numerical_digit += 1
for i in range(1, numerical_digit + 1):
    for str_num in palind(i):
        num = int(str_num)
        if num >= begin and num <= end and prime_number(num):
            print(num)

Code (C-88 points)

#include <stdio.h>
#include <stdlib.h>


long long a,b;
//判断是否是质数 
int pd(long long y) {
    
    
	long long i;
	for(i=2; i*i<=y; ++i) {
    
    
		if(y%i==0) return 0;
	}
	return 1;
}

//判断是否是回文数
int hui(long long n)
{
    
    
	long long num  = n;
	long long m = 0, a;
		while(1){
    
    
			a = n%10;
			m = m*10 + a;
			n = n/10; 
			if( n == 0){
    
    
				break;
			}
		}
	if(m == num)
		return 1;
	else
		return 0; 
	
 } 
 
 
int main()
{
    
    
	scanf("%d %d",&a,&b);
	long long i = a;
	if(a%2 ==0)
		a ++;
	for(i = a;i <= b;i+=2)
	{
    
    

		if(i%3 != 0)
		{
    
    
			int y = hui(i);
			if(y == 1)
			{
    
    
				int x = pd(i);
				if(x ==1)
					printf("%lld\n",i);
			}
		}
		
	}
	return 0;
}

Code (C-100)

#include<stdio.h>
#include<math.h>
int prime(int n) {
    
    //判定素数 
	if(n==1)//特判1 
		return 0;
	if(n%2==0)//2的倍数就回家吧 
		return 0;
	else {
    
    //不然就暴力枚举 
		int i;
		for(i=2; i<=sqrt(n); i++) {
    
    
			if(n%i==0)
				return 0;
		}
		return 1;
	}
}
int hw(int n) {
    
    //判定回文，不懂请参考数字反转 
	int sum=0;
	int k=n;
	while(n!=0) {
    
    
		sum=sum*10+n%10;
		n/=10;
	}
	if(sum==k)//判断是否回文 
		return 1;
	else
		return 0;
}
int main() {
    
    
	int i,n,sum=0,m;
	scanf("%d %d",&n,&m);  //读入两个数 
	for(i=n; i<=m; i++) {
    
    
		if(i==9989900) //如果到了这个数，就break 
			break;
		if(hw(i)&&prime(i))//否则判断是否回文和素数 
			printf("%d\n",i);//输出每个回文质数 
	}
	return 0;//结束程序 
}

12. Xiaoyu is swimming

Question link : https://www.luogu.com.cn/problem/P1423

Question description

Xiaoyu was swimming happily, but she soon found out sadly that she was not strong enough and swimming was very tiring. It is known that Xiaoyu can swim $2$ meters, but as she became more and more tired and her strength became less and less, she could only swim 98% 98\%of the distance of the previous step with each subsequent step.$98\%$ . Now Xiaoyu wants to know, if you want to swim to a distance of$How\; many\; steps\; does\; she\; need\; to\; swim\; in\; a\; place\; of\; s$ meters? Please program to solve this problem.

Input format

Enter a real number $s$ (unit: meter), indicates the target distance to be swam.

Output format

Output an integer, indicating how many steps Xiaoyu needs to swim in total.

Example #1

Sample input #1

4.3

Sample output #1

3

hint

Data guaranteed, $0\le s<100$ , and$s$ has at most one decimal place.

code(python)

s = float(input())

num = 0
step = 2
while s > 0:
    s = s - step
    step = step *0.98
    num += 1
print(num)

Code (C)

#include<stdio.h>
int main()
{
    
    
    float a,b=2,k=2;/*距离可能非整*/
    int i=1;/*由于数据弱就int即可*/
    scanf("%f",&a);/*输入目标*/
    for(i=1;i<=99999999;i++)
    {
    
    if(k>a) break;/*判断是否游到目标*/
     else 
     {
    
    
          b=b*0.98;/*每步能游的距离*/
          k=b+k;/*每步结束后的已游总距离*/
      }}
     printf("%d",i);/*输出步数*/
     return 0;
}

13. [NOIP2011 Popularization Group] Digital Reversal

Question link: https://www.luogu.com.cn/problem/P1307

Question description

Given an integer $N$ , please reverse the digits in each digit of this number to get a new number. The new number should also satisfy the common form of integers, that is, the highest digit of the new number after inversion should not be zero unless the given original number is zero (see Example 2).

Input format

an integer $N$。

Output format

An integer representing the new number after inversion.

Example #1

Sample input #1

123

Sample output #1

321

Example #2

Sample input #2

-380

Sample output #2

-83

hint

【data range】

$-1,000,000,000\leq N\leq $1,000,000,000。

noip2011 popularization group first question

code(python)

num = input()

if num[0] == "-":
    print("-", end="")
    print(int(num[:0:-1]))
else:
    print(int(num[::-1]))

Code (C)

#include <stdio.h>
#include <stdlib.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {
    
    
	int n;
	while(scanf("%d",&n)!=EOF){
    
    
	int m = 0, a;
		while(1){
    
    
			a = n%10;
			m = m*10 + a;
			n = n/10; 
			if( n == 0){
    
    
				break;
			}
		}
		printf("%d\n",m);
	} 
	return 0;
}

14. Money is counted when the moon sets and the crow crows (Fibonacci Sequence)

Question link: https://www.luogu.com.cn/problem/P1720

Question background

(There is no hidden song in this question... No need to guess...)

The final chapter of "The Story of Love and Sorrow Part 1·Heartache".

After eating the pizza, Yueluowudi knew that it was beyond her budget. In order not to show my shame in front of the God of Love and Sorrow, I had no choice but to bite the bullet and count the money...

Question description

After calculating the money, Yueluowu cried and thought: "You fucking tricked me, (hereinafter read in Hokkien) I will depend on my mother and father, (hereinafter read in English) it will be Yiteyou!" So when love and sorrow When the great god asked how much it was, Yueluowudi said a bunch of gibberish. The God of Love and Sorrow said: "Forget it, forget it, I only ask the $How\; much\; does\; n$ type of dishes cost? "The moon fell and the crow cried:

$$F_n=\frac{{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n}{\sqrt{5}}$$

Since the Great God of Love and Sorrow has learned programming, I used 1 1F n F_n$was\; found\; in\; 1$ minute.$F_n$the result of. Yueluowu Tiao was shocked by this. Can you learn love and sorrow? The master asks for $F_n$Is it worth it?

Input format

One natural number $n$。

Output format

only $1$ row of a real number$F_n$, to two decimal places.

Example #1

Sample input #1

6

Sample output #1

8.00

hint

For all data: $0\le n\le 48$。

code(python)

import math



def jia( n ):
    num = float((1 + math.sqrt(5)) / 2)
    sum = 1
    while n > 0:
        sum = float(sum * num)
        n -= 1
    return sum

def jian( n ):
    num = float((1 - math.sqrt(5)) / 2)
    sum = 1
    while n > 0:
        sum = float(sum * num)
        n -= 1
    return sum

if __name__ == '__main__':
    n = int(input())
    a = jia(n)
    b = jian(n)
    num = float((a-b)/math.sqrt(5))
    # print(round(num,3))

    print("{:.2f}".format(num))

Code (C)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
//函数
double  HANSHU(int n){
    
    
	double num = (pow(((1 + sqrt(5.0))/2.0),n) - (pow(((1 - sqrt(5.0))/2.0),n)))/sqrt(5.0);
	return num;
} 
int main(int argc, char *argv[]) {
    
    
	int n;
	while(scanf("%d",&n)!=EOF){
    
    
		double num;
		num = HANSHU(n);
		printf("%.2lf\n",num);
	}
	return 0;
}

15. [Deep Foundation 4. Exercise 5] Find the range/maximum span value

Question link: https://www.luogu.com.cn/problem/P5724

Question description

gives $n$ and$n$ integers$a_i$, please this nnWhat is the range of$n\; integers?$Range means the difference between the maximum value minus the minimum value in a set of numbers.

Input format

The first line enters a positive integer $n$ , represents the number of integers.

Enter $n$ integer$a_1,a_2\dots a_n$, separated by spaces.

Output format

Output an integer representing this $Range\; of\; n$ integers.

Example #1

Sample input #1

6
4 1 5 1 4 1

Sample output #1

4

hint

Data guarantee, $1\le n\le 100$，$0\le a_i\le 1000$。

code(python)

n = int(input())

num = list(input().split(" "))

max1 = int(num[0])
min1 = int(num[0])
for i in range(0,n-1):
    if int(num[i]) > max1:
        max1 = int(num[i])
    if int(num[i]) < min1:
        min1 = int(num[i])

print(max1 - min1)

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>
 
int main()
{
    
    
	int n;
	scanf("%d",&n);
	
	int min = 1000, max = 1;
	
	for(int i = 1;i <= n;i++)
	{
    
    
		int num;
		scanf("%d",&num);
		if(num < min)
		{
    
    
			min = num;
		}
		if(num > max)
		{
    
    
			max = num;
		}
	}
	
	printf("%d",max - min);	
	return 0;	
} 

16. The longest consecutive number

Question link: https://www.luogu.com.cn/problem/P1420

Question description

The input length is $A\; sequence\; of\; positive\; integers\; for\; n$ , requiring the length of the longest consecutive number in the output sequence.

Serial numbers refer to consecutive natural numbers in a sequence from small to large.

Input format

The first line, an integer $n$。

Second line, $n$ integers$a_i$, separated by spaces.

Output format

A number, the number of the longest consecutive numbers.

Example #1

Sample input #1

10
1 5 6 2 3 4 5 6 8 9

Sample output #1

5

hint

Data size and conventions

For $100\%$ data, guaranteed$1\le n\le 10^4$，$1\le a_i\le 10^9$。

code(python)

maxn = 0
n = int(input())
a = list(map(int, input().split()))
lena = len(a)

for i in range(n):
    s = 0
    for j in range(i + 1, n):
        if a[j] == a[j - 1] + 1:
            s += 1
        else:
            break
    maxn = max(maxn, s)

print(maxn + 1)

Code (C)

#include <stdio.h>
#include <stdlib.h>
#define M 10000
/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {
    
    
	int n, i, j, count = 0,sum = 0;
	scanf("%d",&n);
	long long int NUM[M];
	for(i = 0;i < n;i++){
    
    
		scanf("%lld",&NUM[i]);
	}
		for(j = 0;j <= n;j++){
    
    
			if(NUM[j] == NUM[j+1] - 1){
    
    
				count++;
				if(count > sum)
					sum = count;
			}
			else count = 0;		
		}

	printf("%d\n",sum+1);
	return 0;
}

17. [NOIP2012 Popularization Group] Prime Factorization

Question link: https://www.luogu.com.cn/problem/P1075

Question description

Known positive integer $n$ is the product of two different prime numbers, try to find the larger prime number of the two.

Input format

Enter a positive integer $n$。

Output format

Output a positive integer $p$ , which is the larger prime number.

Example #1

Sample input #1

21

Sample output #1

7

hint

$1\le n\le 2\times 10^9$

NOIP 2012 Popularization Group First Question

code(python)

n = int(input())
m = 2
while n % m != 0:
    m += 1
p = int(n / m)
if p <= m:
    p = m
print(p)

Code (C)

#include <stdio.h>

int main()
{
    
    
	int n;
	scanf("%d",&n);
	int m = 2;
	while(n%m != 0)
	{
    
    
		m++;
	}
	int p = n/m;
	if(p <= m) p = m;
	printf("%d",p); 
	return 0;
}

18. [Deep Basis 4. Exercise 8] Find Triangles

Question link: https://www.luogu.com.cn/problem/P5725

Question description

Imitate the example, print out squares in different directions, and then print out a triangular matrix. There is a blank line in the middle.

Input format

The size of the input matrix, not exceeding $9$。

Output format

Output rectangles and squares

Example #1

Sample input #1

4

Sample output #1

01020304
05060708
09101112
13141516

      01
    0203
  040506
07080910

code(python)

cnt = 0

n = int(input())

i = 1

while i <= n * n:
    if i % n == 1 and i != 1:
        print()
        # print("\n")

    if i < 10:
        print("0", end="")

    print(i, end="")

    i += 1

print()
print()
# print("\n\n")

i = 2 * n

while i > 0:
    i -= 2
    for j in range(i):
        print(" ", end="")

    j = 0
    while j < (2 * n - i) // 2:
        cnt += 1
        if cnt < 10:
            print("0", end="")

        print("%d" % cnt, end="")

        j += 1

    print()
    # print("\n")

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>
 
int main()
{
    
    
	int n;
	scanf("%d",&n);
	
	int x = 1; 
	for(int i = 1;i <= n;i++)
	{
    
    
		for(int j = 1;j <= n;j++)
		{
    
    
			if(x < 10)
			{
    
    
				printf("0%d",x);
				x++;
			}
			else
			{
    
    
				printf("%d",x);
				x++;
			}
		}
		printf("\n");
	}
	
	printf("\n");
	int m = 1;
	for(int i = n-1;i >= 0;i--)
	{
    
    
		for(int j = 0;j < i;j++)
		{
    
    
			printf("  ");
		}
		for(int t = i;t <= n-1;t++)
		{
    
    
			if(m < 10)
			{
    
    
				printf("0%d",m);
				m++;
			}
			else 
			{
    
    
				printf("%d",m);
				m++;
			}
		}
		printf("\n");
	}
	return 0;	
} 

19. [Deep Basics 4. Practice 9] Scoring

Question description

Now there are $n(n\le 1000$ judges give scores to the contestants, with scores ranging from$0$$0$ to$10$ . You need to remove the highest score and the lowest score (if there are multiple highest or lowest scores, you only need to remove one), and the average of the remaining ratings is the player's score. Now enter the number of judges and their scores. Please output the final score of the contestants, accurate to$2nd$ place decimal.

Input format

The first line enters a positive integer $n$ , indicating that there is$n$ judges.

Enter $n$ positive integers,$i$ positive integer represents theiithThe score given by$i\; judges.$

Output format

Output one line of two decimals, indicating the player's final score.

Example #1

Sample input #1

5
9 5 6 8 9

Sample output #1

7.67

hint

Data guarantee, $3\le n\le 1000$ , each judge’s score is$0$ to$10$ (including$0$ and$10$ ) an integer between.

code(python)

n = int(input())
grades = list(map(int, input().split()))

new = sorted(grades)
sum = 0
for i in range (1,n-1):
    sum = sum + new[i]
print(float(round(sum/(n-2),2)))

20. [COCI2017-2018#6] Davor

Question link: https://www.luogu.com.cn/problem/P4956

Title translation

After conquering Antarctica, Davor embarked on a new challenge. The next step is an Arctic Circle expedition in Siberia, Greenland, Norway. He will be in 2018 20181212$2018$$December$ 3131Starting on the 31st, we need to raise a total of$nn$$n$ dollars. He plans to raise xxevery Monday$\$\; x$ , raise$x+k$ yuan,..., raised$x+6kyuan$ and raised$52$ weeks. where$x,k$ is a positive integer and satisfies$1\le x\le 100$。

Now please help me calculate $x,When\; k$ is, how much$n$ yuan.

If there are multiple answers, output $x$ is as large as possible,$k$ is as small as possible. Attentionkk$k$ must be greater than$0$。

Question description

After successfully conquering the South Pole, Davor is preparing for new challenges. Next up is the Arctic expedition to Siberia, Greenland and Norway. He begins his travels on 31 December 2018, and needs to collect ​N kunas (Croatian currency) by then. In order to do this, he has decided to put away ​X (​X ≤ 100) kunas every Monday to his travel fund, ​X + K kunas every Tuesday, ​X + 2* ​K every Wednesday, and so on until Sunday, when he will put away ​X + 6* ​K kunas. This way, he will collect money for 52 weeks, starting with 1 January 2018 (Monday) until 30 December 2018 (Sunday).

If we know the amount of money ​N​, output the values ​X and ​K so that it is possible to collect the ​exact money amount in the given timespan. The solution will always exist, and if there are multiple, output the one with the greatest ​X ​ and smallest ​K ​.

Input format

The first line of input contains the integer ​N​ (1456 ≤ ​N​ ≤ 145600), the number from the task.

Output format

The first line of output must contain the value of ​X (​0 < ​X ​≤ 100 ​)​, and the second the value of
K (K ​> 0 ​)​.

Example #1

Sample input #1

1456

Sample output #1

1
1

Example #2

Sample input #2

6188

Sample output #2

14
1

Example #3

Sample input #3

40404

Sample output #3

99
4

code(python)

num  = int(input())

n = 1
k = 1
price = int(num / 364)
n = price - 3 * k
k += 1
while n > 100:
    n = price - 3 * k
    k += 1

print(n)
print(k - 1)

Code (C)

#include<string.h>
#include<stdio.h>
#include<math.h>
#include <stdlib.h>

int main()

{
    
    
	int n, k = 1;
	long long num;
	scanf("%lld",&num);
	
	int price = num / 364;
	do
	{
    
    
		n = price - 3*k;
		k++;
	}
	while(n > 100);
	
	printf("%d\n",n);
	printf("%d\n",k - 1);
	return 0;
}

21. [NOIP2004 Improvement Group] Jinjin’s Savings Plan

Question link: https://www.luogu.com.cn/problem/P1089

Question description

Jinjin has always managed her own pocket money. At the beginning of every month, my mother gives Jinjin $With\; 300$ yuan, Jinjin will budget the expenses for this month, and he can always make the actual expenses consistent with the budget.

In order to let Jinjin learn how to save, her mother proposed that Jinjin can deposit the entire hundred money with her at any time, and at the end of the year she will add $20\%$ is returned to Jinjin. Therefore, Jinjin formulated a savings plan: at the beginning of each month, after receiving the pocket money from her mother, if she expected to have more than 100 100by the end of the month,$100$ yuan or exactly$For\; 100$ yuan, she will deposit the entire hundred with her mother and keep the remaining money in her own hands.

For example $At\; the\; beginning\; of\; November,$ Jinjin still had$83$ yuan, mother gave Jinjin$300$ yuan. Jinjin is expected to be$The\; expenses\; in\; November$ are$180$ yuan, then she will save 200 200with her mother$200$ yuan, keep$183$ yuan. It’s$At\; the\; end\; of\; November$ , Jinjin will have$3$ yuan.

Jinjin discovered that the main risk of this savings plan was that the money deposited with her mother could not be withdrawn before the end of the year. It is possible that at the beginning of a certain month, the money in Jinjin's hands plus the money given by her mother this month is not enough to meet the original budget for this month. If this happens, Jinjin will have to save money and compress the budget this month.

Now please base on $2004$ 1$January$ toDecember 12Use the monthly allowance budget in December to determine whether this situation will occur$.$If not, calculate to$At\; the\; end\; of\; 2004$ , my mother added 20%to the money Jinjin usually saved.$After\; 20\%$ is returned to Jinjin, how much money will Jinjin have in his hands.

Input format

$12$ rows of data, each row contains a value less than$350$ is a non-negative integer, representing$January$ to$December$ Jinjin’s budget.

Output format

an integer. If there is not enough money for a certain month during the implementation of the savings plan, output $-X$，$X$ indicates the first month in which this occurred; otherwise output to2004 2004How much money will Jinjin have in his hands at the end of 2004$?$

Note that Luogu does not need to perform file input and output, but standard input and output.

Example #1

Sample input #1

290
230
280
200
300
170
340
50 
90 
80 
200
60

Sample output #1

-7

Example #2

Sample input #2

290 
230 
280 
200 
300 
170 
330 
50 
90 
80 
200 
60

Sample output #2

1580

code(python)

x = 0
y = 0
save = 0
flag = 1
month = 0
for i in range(1, 13):
    x += 300
    y = int(input())
    x -= y
    if (x < 0):
        flag = 0
        month = i
        break
    save += int(x / 100)
    x = x % 100
if (flag == 1):
    print(120 * save + x)
else:
    print(-month)

Code (C)

#include <stdio.h>
int main()
{
    
    
    int n;
    int sum = 0;	        //初始化，用于记录每个月剩余有多少钱
    int sum0 = 0;	        //如果中间没有出现缺钱的情况，那么就用这个sum0来存储每个月在妈妈手里有多少钞票
    for (int i = 1; i <= 12; i++) //一年十二个月，不多做解释了
    {
    
    
        sum += 300;     //妈妈给钱
        scanf("%d", &n);
        sum = sum - n;
        if (sum < 0)     //如果消费大于开支则为负数，直接输出ok
        {
    
    
            printf("-%d\n", i); //特别注意负号
            return 0;
        }
        else
            sum0 += sum / 100 * 100; //如果不是就往妈妈手里存钱
        sum = sum - sum / 100 * 100; //每个孩子手里剩余多少钱
    }
    printf("%d\n", sum0 * 12 / 10 + sum); //如果上述未有输出，那么就可以打印输出多少钱了
    return 0;
}