HDU - 1028 Ignatius and the Princess III 生成函数

int c1[130];
int c2[130];

int main() {
    int n;
    while (~scanf("%d", &n)) {
        for (int i = 0; i <= n; i++) c1[i] = 1, c2[i] = 0;
        for (int i = 2; i <= n; i++) {
            for (int j = 0; j <= n; j++)
                for (int k = 0; k + j <= n; k += i)
                    c2[k + j] += c1[j];
            for (int j = 0; j <= n; j++) {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        printf("%d\n", c1[n]);
    }
}