1159: 偶数求和 简单易懂的优质题解

解题思路:巧用循环

注意事项:

参考代码:

#include<iostream>

using namespace std;

int main()

{

int n = 0, m = 0 ,g;

while(cin >> n >> m)

{ 

int c = n / m，aver = 0, k = 1, sum = 0;

for (int i = 0; i < c; i++)

{

for (int j = 0; j < m; j++)

{

sum += 2 * k;

++k;

}

aver = sum / m;

cout << aver << " ";

sum = aver = 0;

}

if (n - (c * m) > 0)

{

for (g = 0; g < n - (c * m); g++)

{

sum += 2 * k;

++k;

}

aver = sum / g;

cout << aver << " ";

g = 0;

}

}

return 0;

}