Problem C. 欧皇 ————2019.10.12

题目：

 

 

 

 

 

 

 

 

 再次感激土蛋

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
const ll mod = 1e9 + 7;
ll C[1005][1005];
 
void pre(){
    C[0][0] = 1;
    for(int i = 1; i <= 1000; i ++){
        C[i][0] = 1;
        for(int j = 1; j <= i; j ++){
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
        }
    }
}
int n, m, c;
int a[105];
ll f[11][31][31];
ll g[11][31][31], h[11][31][31];
int main(){
    freopen("europe.in", "r", stdin);
    freopen("europe.out", "w", stdout);
    
    scanf("%d%d%d", &n, &m, &c);
    for(int i = 1; i <= c; i ++){
        scanf("%d", &a[i]);
    }   
    pre();
    for(int k = 1; k <= c; k ++){
        for(int i = 1; i <= n; i ++){
            for(int j = 1; j <= m; j ++){
                h[k][i][j] = C[i * j][a[k]];
                ll flag = -1;
                for(int jj = j - 1; jj >= 1; jj --){
                    h[k][i][j] = (h[k][i][j] + C[j][jj] * C[i * jj][a[k]] % mod * flag + mod) % mod;
                    flag = -flag;
                }
            }
        }
    }
    for(int k = 1; k <= c; k ++){
        for(int j = 1; j <= m; j ++){
            for(int i = 1; i <= n; i ++){
                g[k][i][j] = h[k][i][j];
            //  printf("%d %d %d %lld
", k, i, j, g[k][i][j]);
                ll flag = -1;
                for(int ii = i - 1; ii >= 1; ii --){
                    g[k][i][j] = (g[k][i][j] + C[i][ii] * h[k][ii][j] % mod * flag + mod) % mod;
                    flag = -flag;
                }
            //  printf("%d %d %d %lld
", k, i, j, g[k][i][j]);
            }
        }
    }
    f[0][0][0] = 1;
    for(int k = 1; k <= c; k ++){
        for(int i = 1; i <= n; i ++){
            for(int j = 1; j <= m; j ++){
                for(int ii = 0; ii <= i - 1; ii ++){
                    for(int jj = 0; jj <= j - 1; jj ++){
                        if((i - ii) * (j - jj) < a[k]) continue;
                        f[k][i][j] = (f[k][i][j] + C[i][i - ii] * C[j][j - jj] % mod * f[k - 1][ii][jj] % mod * g[k][i - ii][j - jj]) % mod;
                    }
                }
            }
        }
    }
    ll ans = 0;
    for(int i = 1; i <= n; i ++){
        for(int j = 1; j <= m; j ++){
            ans = (ans + C[n][i] * C[m][j] % mod * f[c][i][j]) % mod;
        }
    }
    //printf("%d %d %d
", n, m, c); 
    printf("%lld
", ans);
    return 0;
}