UVa 11806 - Cheerleaders (组合计数+容斥原理)

《训练指南》p.108

#include <cstdio>
#include <cstring>
#include <cstdlib>

using namespace std;

const int MOD = 1000007;

const int MAXN = 500;

int C[MAXN][MAXN];

void init()
{
    memset( C, 0, sizeof(C) );
    C[0][0] = 1;
    for ( int i = 0; i < MAXN; ++i )
    {
        C[i][0] = C[i][i] = 1;
        for ( int j = 1; j < i; ++j )
        C[i][j] = ( C[i-1][j] + C[i-1][j-1] ) % MOD;
    }
    return;
}

int main()
{
    init();
    int T, cas = 0;
    scanf( "%d", &T );
    while ( T-- )
    {
        int M, N, K;
        int ans = 0;
        scanf( "%d%d%d", &M, &N, &K );
        for ( int S = 0; S < ( 1 << 4 ); ++S )
        {
            int cnt = 0;
            int r = M, c = N;
            if ( S & 1 ) --r, ++cnt;
            if ( S & 2 ) --r, ++cnt;
            if ( S & 4 ) --c, ++cnt;
            if ( S & 8 ) --c, ++cnt;
            if ( cnt & 1 ) ans = ( ans + MOD - C[r*c][K] )%MOD;
            else ans = ( ans + C[r*c][K] )%MOD;
        }
        printf( "Case %d: %d\n", ++cas, ans );
    }
    return 0;
}