洛谷 P1373 小a和uim之大逃离

题目传送门

dp,f[i][j][k][1/0]表示到(i,j)uim/小a魔瓶里的液体为k(经取模后)的方案数

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int n,m,k,a[801][801],o;
int f[801][801][20][2];
long long ans;

int main() {
    scanf("%d%d%d",&n,&m,&k);
    k++;
    for(int i = 1;i <= n; i++)
        for(int j = 1;j <= m; j++) {
            scanf("%d",&a[i][j]);
            f[i][j][a[i][j]%k][0] = 1;
        }
    for(int i = 1;i <= n; i++)
        for(int j = 1;j <= m; j++)
            for(int p = 0;p < k; p++) {
                f[i][j][p][0] = (f[i][j][p][0] + f[i-1][j][(p-a[i][j]+k)%k][1]) % 1000000007;
                f[i][j][p][0] = (f[i][j][p][0] + f[i][j-1][(p-a[i][j]+k)%k][1]) % 1000000007;
                f[i][j][p][1] = (f[i][j][p][1] + f[i-1][j][(p+a[i][j])%k][0]) % 1000000007;
                f[i][j][p][1] = (f[i][j][p][1] + f[i][j-1][(p+a[i][j])%k][0]) % 1000000007;
            }
    for(int i = 1;i <= n; i++)
        for(int j = 1;j <= m; j++)
            ans = (ans + f[i][j][0][1]) % 1000000007;
    printf("%lld",ans);
    return 0;
}