Luogu - Mathematics 1 (good and wrong questions)

[NOIP2000 Improvement Group] Base conversion

topic description

We can use this way to represent a decimal number: multiply each Arabic numeral by an exponent of the position of the number, with 10 10Form of the sum of powers of base$10\; .$eg$123$ can be expressed as$1\times 10^2+2\times 10^1+3\times 10^0$ such a form.

Similarly, for binary numbers, it can also be expressed as each binary number multiplied by an exponent based on the position of the number, with 2 2Form of the sum of powers of base$2\; .$

Generally speaking, any positive integer $R$ or a negative integer$-R$ can be chosen as the base of a number system. If by$R$ or$-R$ is the base number, the numbers to be used are$0,1,....R-1$。

For example when $R=At\; 7$ , the required numbers are$0,1,2,3,4,5,6$ , which is rather$R$ or$-R$ is irrelevant. If the absolute value of the base number exceeds$10$ , in order to represent these numbers, English letters are usually used to represent those greater than$9$ digits. For example for$For\; hexadecimal$ numbers, use$A$ means$10$ , use$B$ means$11$ , with$C$ means$12$ , and so on.

In negative base numbers, $-R $ is used as the base, such as $-15$ (decimal) is equivalent to$(110001{)}_{-2}$ （$-binary$ ), and it can be expressed as$Sums\; of\; power-of\; -2$ series:

$$(110001{)}_{-2}=1\times (-2{)}^5+1\times (-2{)}^4+0\times (-2{)}^3+0\times (-2{)}^2+0\times (-2{)}^1+1\times (-2{)}^0$$

Design a program that reads in a base of a decimal number and a negative base, and converts the decimal number into a number in this negative base.

input format

Each line of input has two input data.

The first is the decimal number $n$ .
The second is the base of negative base numbers$-R$。

output format

Output this negative base number and its base, if the base exceeds $10$ , then refer to$Hexadecimal$ way processing.

Example #1

Sample Input #1

30000 -2

Sample output #1

30000=11011010101110000(base-2)

Example #2

Sample Input #2

-20000 -2

Sample output #2

-20000=1111011000100000(base-2)

Example #3

Sample Input #3

28800 -16

Sample output #3

28800=19180(base-16)

Example #4

Sample Input #4

-25000 -16

Sample output #4

-25000=7FB8(base-16)

hint

【Data range】
for $100\%$ of the data,$-20\le R\le -2$，$|n|\le 37336$。

The first question of NOIp2000 improvement group

train of thought

First of all, we need to know how the decimal system is converted into other base systems. Suppose the decimal number 15 is converted to binary. $15=((1\times 2+1)\times 2+1)\times 2+1$ , only need to continue to take the modulo, divisible. The loop goes on until the divisibility is 0, the modulus is arranged1111. In essence, the number of digits obtained for each digit represents the number of decimal weights of this number, that is, multiplied by (number of digits - 1) 2.
For negative decimal numbers, it is assumed to be$-Binary$ can be expressed as$15=(((1\times -2)\times -2)\times -2)\times -2-1$ , according to the original method, the modulus has a negative number. It needs to be transformed according to$dividend=business\times divisor+remainder\equiv (Business+1)\times divisor+remainder-Divisor$ ,$15=(((1\times -2)\times -2)\times -2+1)\times -2+1$ , the obtained -binary number is10011. Then in the transformation, if a negative remainder is encountered, the resulting$remainder-Divisor$ ,$business+1$ is enough.

the code

def convert(n, base) :
	res = ''
	if n == 0 :
		return 0
	while n :
		t = n % base
		n //= base
		if t < 0 :
			n += 1
			t -= base
		if t >= 10 :
			t = chr(ord('A') + t - 10)
		res = str(t) + res
	return res
n, base = map(int, input().split())
res = convert(n, base)
print(f"{
      
      n}={
      
      res}(base{
      
      base})")

Number of straight line intersections

topic description

Suppose there are $If\; there\; are\; N$ straight lines and no three lines have the same point, how many different intersections can these straight lines have in total?

input format

one line, an integer $N$ , stands for$N$ straight lines.

output format

One line, an integer, indicates the total number of schemes.

Example #1

Sample Input #1

4

Sample output #1

5

hint

For all data, satisfy $1\le N\le 25$。

train of thought

For a pile (N) of straight lines, it can be divided into two piles, one pile § is the straight lines parallel to each other, and one pile (N -r) is the straight lines not parallel to the other pile. The point of intersection between these two lines is $p\times (N-r)$ . There is no point of intersection between parallel lines, and the point of intersection can also be calculated according to this method for straight lines that are not parallel to another group, until the number of parallel lines is 0 and there is no point of intersection.
Recursive thinking, assuming that there are currently n straight lines, enumerate the number of parallel straight lines, record the intersection points between two sets of straight lines, and continue to recursively process the non-parallel straight lines until the remaining straight lines are 0.

the code

N = 25 * 25 + 10
st = [False] * N
ans = 0
def dfs(n, s) : # 分别表示当前处理的直线堆、已算出的交点数
	global ans
	if n == 0 : # 终止条件
		if not st[s] :
			ans += 1
			st[s] = True
		return
	# 枚举平行线数
	for i in range(1, n + 1) :
		dfs(n - i, s + i * (n - i))
n = int(input())
dfs(n, 0)
print(ans)

coding

topic description

Encoding is often applied to ciphertext or compressed transmission. Here we use one of the simplest encoding methods for encoding: encode some regular words into numbers.

There are 26 letters {a, b, ..., z} in the alphabet, these special words are not more than 6 in length and the letters are arranged in ascending order. Putting all such words together, in lexicographical order, a word's encoding corresponds to its position in the dictionary.

For example:

a→1 b→2 z→26 ab→27 ac→28

Your task is to find the code for the given word.

input format

Just one line, encoded words.

output format

Only one line, corresponding to the encoding. If the word is not in the alphabet, output 0.

Example #1

Sample Input #1

ab

Sample output #1

27

train of thought

Take cgxas an example

1. First of all, a word with a length of 1 must be smaller than it $C_{26}^1$
2. A word of length 2 must be smaller than it $C_{26}^2$
3. The case where the first letter is less cthan must be less than it $C_{24}^2$
4. The case where the first letter is equal c, and gxthe word less than it must be smaller than it (recursive processing is enough).

the code

It's Graduation Season II

topic background

"Dingling, ringing", as the bell of the last subject of the college entrance examination rang, three years of youth suddenly froze at this moment. The joy of graduation is no match for the reluctance of parting, looking forward to the future and still not forgetting the past songs. The laughter and tears of more than a thousand days and nights were all gathered at the graduation party. I believe that this must be the most unforgettable moment in my life!

topic description

There was a rehearsal once, and the teacher was not very satisfied. Of course, it is obviously unreasonable to take the number of each student to find the greatest common divisor. So the teacher rated each student with an ability value. So now the problem becomes, from nnPick kk$out\; of\; n$ students$K$ individuals maximize their tacit understanding (that is, the greatest common divisor of ability values). But because there are too many programs, and the number of people needed for each program is unknown. The teacher wants to know what is the maximum level of rapport that can be achieved in all situations. It's even more troublesome now, so I'll leave it to you~

PS: The greatest common divisor of a number is itself.

input format

The first line is a positive integer $n$。

The second line is nnA positive integer separated by$n\; spaces,\; representing\; the\; ability\; value\; of\; each\; student.$

output format

Total nnline$n$$i$ act$k=$The maximum degree of tacit understanding in case$i\; .$

Example #1

Sample Input #1

4
1 2 3 4

Sample output #1

4
2
1
1

hint

[source of topic]

lzn original

【data range】

Note that the maximum value of the ability value in the input data is $\text{inf}$。

• For $20\%$ of the data,$n\le 5$，$\text{inf}\le 10^3$；
• For the other $30\%$ of the data,$n\le 100$，$\text{inf}\le 10$；
• For $100\%$ of the data,$n\le 10^4$，$\text{inf}\le 10^6$。

train of thought

Solve 1 − $1-$The greatest common factor of$n\; numbers.$A violent idea is to use the number of combinations from 1 to n, and then find the data that the greatest common factor can pass 30%.
If you want to ac, you have to start from the factor itself. Suppose you want to find the greatest common factor of k numbers, then for this factor, it must be shared by k numbers or more. Then we can compare the factors by counting the number of factors of all numbers. The greatest factor of a number greater than or equal to k is the greatest common factor of k numbers.

the code

N = 1000010

C = [0] * N
A = [0] * N
t = 0

def get_factor(num) : # 求约数(O(sqrt(n)))
	i = 1
	while i <= num // i :
		if num % i == 0 :
			C[i] += 1
			if i != num // i :
				C[num // i] += 1
		i += 1

n = int(input())
A[1 : n + 1] = list(map(int, input().split()))
# 记录所有数因子情况
for i in range(1, n + 1) :
	t = max(t, A[i])
	get_facotor(A[i])

for i in range(1, n + 1) : # 因子数也具有单调性
	if C[t] < i : t-= 1
	print(t)

check-in question

topic background

This is a check-in question!

It is recommended to read the data range carefully before doing the question!

topic description

We define a function: $\mathrm{qiandao}\left(x\right)$ is less than or equal toxxAmong the numbers of$x$$x$ is the number of numbers thatare not coprime

This question is used as a check-in question, given $l$ and$r$ , find:

$$\sum _{i=l}^r\mathrm{qiandao}\left(i\right)\mathrm{mod}666623333$$

input format

A line of two integers, $l$、$r$。

output format

One integer per line represents the answer.

Example #1

Sample Input #1

233 2333

Sample output #1

1056499

Example #2

Sample Input #2

2333333333 2333666666

Sample output #2

153096296

hint

• For $30\%$ of the data,$l,r\le 10^3$。
• For $60\%$ of the data,$l,r\le 10^7$。
• For $100\%$ of the data,$1\le l\le r\le 10^{12}$，$r-l\le 10^6$。

(30 points)

'''
筛法筛欧拉函数，用筛好的欧拉函数，继续筛区间欧拉函数
'''
from math import ceil
N = 1000010
MOD = 666623333

eula = [0] * N
st = [False] * N
primes = []

def init(n) :
    for i in range(2, n + 1) :
        if not st[i] :
            primes.append(i)
        for j in primes :
            if i * j > n : break
            st[i * j] = True
            if i % j == 0 :
                break

init(N - 1)
l, r = map(int, input().split())

def get_eula(x) :
    res = x
    for i in primes :
        if i > x : break
        flag = True
        while x % i == 0 :
            x //= i
            flag = False
        if not flag :
            res = res * (i - 1) // i
    return res

# 区间筛质数
st = [False] * N
for i in primes[:-1] :
    if i > r : break
    j = max(2, ceil(l / i)) * i
    while j <= r :
        st[j - l] = True
        j += i

for i in range(l, r + 1) :
    if i >= 2 and not st[i - l] :
        eula[i - l] = i - 1
    elif i < 2 : eula[i] = 1
    else :
        eula[i - l] = get_eula(i)
res = 0
for i in range(l, r + 1) :
    res = (res + i - eula[i - l]) % MOD
print(res)