UVA 11806 组合数学+容斥

UVA: https://vjudge.net/problem/UVA-11806

AC代码

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("\n")
#define debug(a,b) cout<<a<<" "<<b<<" "<<endl
#define ffread(a) fastIO::read(a)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
//const int maxn=1e4+10;
const ll mod=1e6+7;
const int maxn = 1010;
int C[maxn][maxn];
int n, m, k;

void init()
{
    C[0][0] = 1;
    for (int i = 1; i < maxn; 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 main()
{
    int t,kase=1;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&m,&n,&k);
        int ans=C[n*m][k];
        ans=(ans-2*C[(m-1)*n][k]%mod+mod)%mod;
        ans=(ans-2*C[(n-1)*m][k]%mod+mod)%mod;
        ans=(ans+4*C[(n-1)*(m-1)][k])%mod;
        ans=(ans+C[(n-2)*m][k])%mod;
        ans=(ans+C[(m-2)*n][k])%mod;
        ans=(ans-2*C[(n-2)*(m-1)][k]%mod+mod)%mod;
        ans=(ans-2*C[(m-2)*(n-1)][k]%mod+mod)%mod;
        ans=(ans+C[(m-2)*(n-2)][k])%mod;
        printf("Case %d: %d\n",kase++,ans);
    }
}