LightOJ - 1125 Divisible Group Sums(DP)

题目大意：给你N个数，从中挑选M个，使得M个数的和能被S整除

解题思路：用dp[i][j][k][mod]表示前i个数中选出了j个，除数为k，余数为mod的有多少种情况，接着就可以转移了，分两种情况，选和不选进行转移

这里有个坑点，就是给的数有可能是负数，所以要先转正

#include <cstdio>
#include <cstring>
typedef long long LL;
const int N = 210;

//dp[i][j][k][l],表示从前i个种，选择j个，除数为k，余数为l的个数
LL val
, dp
[10][21][20], num
[22];
int n, q, cas = 1;

void init() {
scanf("%d%d", &n, &q);
for (int i = 1; i <= n; i++)
scanf("%lld", &val[i]);
for (int i = 1; i <= n; i++)
for (long long j = 1; j <= 20; j++)
num[i][j] = (val[i] + j * 100000000000) % j;

memset(dp, 0, sizeof(dp));

for (int k = 1; k <= 20; k++)
dp[1][1][k][num[1][k]] = 1;

for (int j = 2; j <= n; j++)
for (int k = 1; k <= 20; k++) {
dp[j][1][k][num[j][k]]++;
for (int l = 0; l < 20; l++)
dp[j][1][k][l] += dp[j - 1][1][k][l];
}

for (int i = 2; i <= 10; i++)
for (int j = i; j <= n; j++)
for (int k = 1; k <= 20; k++)
for (int l = 0; l < 20; l++) {
//选
dp[j][i][k][(num[j][k] + l) % k] += dp[j - 1][i - 1][k][l];
//不选
dp[j][i][k][l] += dp[j - 1][i][k][l];
}
}

void solve() {
int d, m;
printf("Case %d:\n", cas++);
while (q--) {
scanf("%d%d", &d, &m);
printf("%lld\n", dp
[m][d][0]);
}
}

int main() {
int test;
scanf("%d", &test);
while (test--) {
init();
solve();
}
return 0;
}