HDU2015 偶数求和

#include <stdio.h>#include <string.h>intmain(int argc, char **argv) {    int n, m, i;    int num[105];    int res[105];    int sum, cnt;    for (i = 1; i < 105; ++ i) {        num[i] = i * 2;    }    while (scanf("%d%d", &n, &m) == 2) {            sum = 0;            cnt = 0;            memset(res, 0, sizeof(res));            for (i = 1; i <= n; ++ i) {                sum += num[i];                if (i % m == 0) {                        res[cnt++] = sum / m;                        sum = 0;                }            }            if (n > n / m * m) {                    for (i = 0; i < cnt; ++ i)                        printf("%d ", res[i]);                printf("%d\n", sum / (n - n / m * m));            }            else {                for (i = 0; i < cnt - 1; ++ i)                        printf("%d ", res[i]);                printf("%d\n", res[cnt - 1]);            }    }    return 0;}