UVA11806 Cheerleaders

Cheerleaders

大意：n * m个网格中放若干石子，要求最上面一行、最下面一行、最左边一列、最右边一列必须有石子

没有石子比有石子好求，考虑容斥

小集合是最上面一行没有、最下面一行没有、最左边一列没有、最右边一列没有

注意MOD不是质数！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(long long &a, long long &b)
{
    long long tmp = a;a = b;b = tmp;
}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const long long INF = 0x3f3f3f3f;
const long long MAXK = 500;
const long long MAXN = 20;
const long long MOD = 1000007;

long long n,m,k,t,C[MAXN * MAXN + 5][MAXK + 5];

int main()
{
    read(t);long long cnt = 0;
    C[0][0] = 1;
    for(register int i = 0;i <= MAXN * MAXN;++ i)
    {
        C[i][0] = 1;C[i][i] = 1;
        for(register int j = 1;j <= i;++ j)
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
    }
    for(;t;--t)
    {
        ++ cnt;
        read(n), read(m), read(k);
        long long ans = 0;
        for(long long S = 0;S < 16;++ S)
        {
            long long num = 0, a = n, b = m;
            if(S & 1) -- a, ++ num;
            if(S & 2) -- b, ++ num;
            if(S & 4) -- a, ++ num;
            if(S & 8) -- b, ++ num;
            if(num & 1) ans = (ans - C[a* b][k] + MOD) % MOD;
            else ans = (ans + C[a * b][k]) % MOD;
        }
        printf("Case %lld: %lld
", cnt, ans);
    }
    return 0;
}