UVa 11806 Cheerleaders (数论容斥原理)

题意：给定一个n*m的棋盘，要放k个石子，要求第一行，最后一行，第一列，最后一列都有石子，问有多少种放法。

析：容斥原理，集合A是第一行没有石子，集合B是最后一行没有石子，集合C是第一列没有石子，集合D是最后一列没有石子，如果某一行或某一列，

没有，那么就相当于减少一行或者一列。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <ctime>
#include <cstdlib>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 500 + 5;
const LL mod = 1000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL C[maxn][maxn];
 
void init(){
    for(int i = 0; i < maxn; ++i)  C[i][i] = C[i][0] = 1;
    for(int i = 1; i < maxn; ++i)
        for(int j = 1; j <= i; ++j)
            C[i][j] = (C[i-1][j] + C[i-1][j-1]) % mod;
}
 
int main(){
    init();
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        int k;
        scanf("%d %d %d", &n, &m, &k);
        int ans = 0;
        for(int i = 0; i < 16; ++i){
            bool ok = true;
            int r = n, c = m;
            if(i & 1){ ok = !ok; --r; }
            if(i & 2){ ok = !ok; --r; }
            if(i & 4){ ok = !ok; --c; }
            if(i & 8){ ok = !ok; --c; }
            if(ok)  ans = (ans + C[r*c][k]) % mod;
            else ans = (ans - C[r*c][k] + mod) % mod;
        }
        printf("Case %d: %d\n", kase, ans);
    }
    return 0;
}