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;

}

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UVa 11806 Cheerleaders (数论容斥原理)

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