[Educoder homework] C&C++ control structure training

[Educoder homework] C&C++ control structure training

Logically speaking, after completing this course and learning a function and recursion, you can participate$WorldFinal$了…

T1 branch structure: is it a leap year?

It’s very simple, there are only two types of leap years, $400$'s multiple$4$Multiple but not necessary$Multiple\; of\; 100$.

// 包含两种I/O库，可以使用任一种输入输出方式
#include <stdio.h>
#include <iostream>
using namespace std;

int main()
{
    
    
    int year;
    // 请在此添加代码，判断输入的年份是否位闰年，是则输出"Yes"，否则输出"No"
    /********** Begin *********/
	cin >> year;
	if ((year % 400 == 0) || (year % 4 == 0 && year % 100 != 0)) puts("Yes");
	else puts("No");
    
    
    /********** End **********/
    return 0;
}

T2 branch structure: day of year

As for , we can handle it a little more cleverly. The general idea is to add up the number of days before this month and then add what day of the month it is now.
for $2$ month processing, we can use a $t$ to make the operation universal. For example, leap year$t$ i.e.$1$，No 则$t$ i.e.$0$, so there is no need to judge again when adding.

// 包含两种I/O库，可以使用任一种输入输出方式
#include <stdio.h>
#include <iostream>
using namespace std;

int main()
{
    
    
    // y-年，m-月,d-日，n-第几天
    int y, m, d, n;
    // 请在此添加代码，计算并输出指定日期是第几天
    /********** Begin *********/
	cin >> y >> m >> d ;
	int t = (y % 400 == 0) || (y % 4 == 0 && y % 100 != 0);
	n = 0;
	for (int i = 1; i < m; i ++ ) {
    
    
		if (i == 2) n += 28 + t;
		else if (i == 4 || i == 6 || i == 9 || i == 11) n += 30;
		else n += 31;
	}
	n += d;
    
    
    /********** End **********/
    printf("%d-%d-%d是第%d天\n",y,m,d,n);
    return 0;
}

T3 branch structure: maximum number of rearrangements

Ahem, I used one$sort$, otherwise there are six judgments, a total of $3!$Special circumstances.

// 包含两种I/O库，可以使用任一种输入输出方式
#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
    
    
    // n-输入的数，m-重排后的数
    int n, m;
    // 请在此添加代码，输入一个小于1000的正整数，重排出最大的数存入m中
    /********** Begin *********/
	cin >> n ;
	int a[3];
	a[0] = n / 100;
	a[1] = n / 10 - a[0] * 10;
	a[2] = n % 10;
	sort(a, a + 3);
	m = a[2] * 100 + a[1] * 10 + a[0];
    
    /********** End **********/
    // 输出重排后的数
    cout << m << endl;
    return 0;
}

T4 Loop Structure: Black Hole Trap

No different from the previous question, just need to use $while$ loop. Just pay attention to the exit conditions and counters.

// 包含两种I/O库，可以使用任一种输入输出方式
#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
    
    
    int n;
    // 请在此添加代码，输出整数进入黑洞过程
    /********** Begin *********/
	cin >> n ;
	int t = 1;
	while (n != 495) {
    
    
		int mdl = n, a[3];
		a[0] = mdl / 100;
		a[1] = mdl / 10 - a[0] * 10;
		a[2] = mdl % 10;
		sort(a, a + 3);
		printf("%d:%d-%d=%d\n", t, a[2] * 100 + a[1] * 10 + a[0], a[0] * 100 + a[1] * 10 + a[2], (a[2] - a[0]) * 99);
		n = (a[2] - a[0]) * 99;
		t ++ ;
	}
    
    /********** End **********/
    return 0;
}

T5 cyclic structure: is it a prime number?

Just use the most efficient method, starting from$2$to$n-1$, if there is one that can divide it, it is not a prime number.
Optimize it slightly, for example, just start from $2$Judgment reached$\sqrt{n}$ is enough, because if there is one greater than $\sqrt{n}$target number$x$满足$\exists y$使得$xy=n$，nan么$and=\frac{n}{x}$must be less than$\sqrt{n}$, that is, it can be judged.
Of course, there are also quite fast algorithms $Millar-Rabin$. Simply put, it is the fastest known algorithm for determining prime numbers based on binary fast multiplication and Fermat's little theorem. But $The\; MR$ algorithm is based on random numbers, so its constants may affect accuracy and are difficult for beginners to understand, but I will introduce it in detail.

// 包含两种I/O库，可以使用任一种输入输出方式
#include <stdio.h>
#include <iostream>
using namespace std;

int main()
{
    
    
    int n;

    // 请在此添加代码，输入正整数n，如果n是素数则输出“Yes”，否则输出“No”
    /********** Begin *********/
	cin >> n;
	if (n == 1) {
    
    
		puts("No");
		return 0;
	}
	if (n == 2) {
    
    
		puts("Yes");
		return 0;
	}
	for (int i = 2; i < n; i ++ ) {
    
    
		if (n % i == 0) {
    
    
			puts("No");
			return 0;
		}
	}
	puts("Yes");
    /********** End **********/

    return 0;
}

T6 loop structure: sum of prime numbers

It's nothing more, it's just a cycle.

// 包含两种I/O库，可以使用任一种输入输出方式
#include <stdio.h>
#include <iostream>
using namespace std;

int main()
{
    
    
    int n, k;
    // 请在此添加代码，输入n和k，并输出n以内k个素数以及它们的和
    /********** Begin *********/
	cin >> n >> k;
	int sum = 0;
	for (int i = n; i; i -- ) {
    
    
		bool flag = true;
		for (int j = 2; j < i; j ++ ) if (i % j == 0) flag = false;
		if (flag && i != 1) {
    
    
			sum += i;
			cout << i << ' ' ;
			k -- ;
			if (!k) {
    
    
				cout << sum << endl ;
				return 0;
			}
		}
	}
	cout << sum << endl ;
	return 0;

    /********** End **********/

    return 0;
}