#include<stdio.h>
#include<stdlib.h>
int main(void) {
int a[101], i, j, t = 0;
int sum = 0;
int n, m;
int k = 1;
int b[101];
while (scanf_s("%d%d", &n, &m) != EOF) {
for (i = 1; i <= 100; i++) {
a[i] = 2 * i;
}
for (j = 1; j <= n; j++) {
sum += a[j];
if (j % m == 0 && m * k <= n) {
b[k] = sum / m;
sum = 0;
k++;
}
else {
if (m * k > n) {
t++;
if (j == n) {
b[k] = sum / t;
k++;
sum = 0;
}
}
}
}
for (i = 1; i < k-1; i++) {
printf("%d ", b[i]);
}
printf("%d\n", b[i]);
k = 1;
t = 0;
}
}
杭电偶数求和
最新推荐文章于 2021-11-19 12:38:28 发布