NEFUOJ 424偶数求和

#include<stdio.h>
#include<stdlib.h>
int main()
{
	int n, m,sum,a0,isprime;
	while (scanf("%d%d", &n, &m) != EOF)
	{
		a0 = 2,sum = 0;
		isprime = 0;
		for (int i = 1; i <= n; i++)
		{
			sum += a0;
			a0 = a0 + 2;				
			if (i % m == 0)
			{
				if (isprime == 1)printf(" ");
				printf("%d", sum / m);       //   每m个一组
				sum = 0;				     //输出一次 sum清空
				isprime = 1;
			}
		}
		if (n % m == 0)printf("\n");		 //n是m的整数倍结束
		else printf(" %d\n", sum / (n % m)); //剩余的再输出一下
	}
	return 0;
}

标签：main,int,sum,偶数,a0,NEFUOJ,isprime,include,424
来源： https://blog.csdn.net/weixin_62486861/article/details/121718837