Uva 11806 Cheerleaders

题目链接:

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2906

本题我本来想逐步分类四个角的情况。但是发现分类之间是没有重复的，而内部是有重复的。错误代码如下。

#include <stdio.h>#include <stdlib.h>#include <algorithm>#include <string.h>#include <math.h>#include <iostream>using namespace std;#define MOD 1000007#define Maxn 405int c[Maxn+10][Maxn+10];void init(){    memset(c,0,sizeof(c));    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;        }    }}int solve(int m,int n,int k){    int ans = 0;    int total = m * n;    //不可能    printf("ans1 = %d\n",ans);    if(k > total) return ans;    //无角放棋    if(k >= 4 && total >= 8 && m>2 && n>2)    {        if((k-4)<=(total-8))        {            ans = (ans + (m-2)*(m-2)*(n-2)*(n-2)*c[total-8][k-4])%MOD;        }    }    printf("ans2 = %d\n",ans);    //只有一角放棋    if(n>2 && m>2 && k>=3 && total>=6)    {        if((k-3)<=(total-4))        {            ans = (ans + 4*(m-2)*(m-2)*c[total-6][k-3])%MOD;        }    }    printf("ans3 = %d\n",ans);    //只有二角放棋    if(1)    {        //对角线        if(total>=4 && k>=2 && (k-2)<=(total-4))        {            ans = (ans + 2*c[total-4][k-2])%MOD;        }        printf("ans4 = %d\n",ans);        //一行        if(n>2 && k>=3 && total>=5 && (k-3)<=(total-5))        {            ans = (ans + 2*(n-2)*c[total-5][k-3])%MOD;        }        printf("ans5 = %d\n",ans);        //一列        if(m>2 && k>=3 && total>=5 && (k-3)<=(total-5))        {            ans = (ans + 2*(m-2)*c[total-5][k-3])%MOD;        }        printf("ans6 = %d\n",ans);    }    //只有三角放棋    if(total>=4 && k>=3 && (k-3)<=(total-4))    {        ans = (ans + 4 * c[total-4][k-3])%MOD;    }    printf("ans7 = %d\n",ans);    //四角都放棋    if(total>=4 && k>=4 && (k-4)<=(total-4))    {        ans = (ans + c[total-4][k-4])%MOD;    }    printf("ans8 = %d\n",ans);    return ans;}int main(){    #ifndef ONLINE_JUDGE        freopen("../in.txt","r",stdin);    #endif    int t;    int cas = 0;    int ans;    int m,n,k;    init();    scanf(" %d",&t);    while(t--)    {        cas++;        scanf(" %d %d %d",&m,&n,&k);        ans = solve(m,n,k);        printf("Case %d: %d\n",cas,ans);    }    return 0;}

正确的方法是使用容斥原理，代码如下：

#include <stdio.h>#include <stdlib.h>#include <algorithm>#include <string.h>#include <math.h>#include <iostream>using namespace std;#define MOD 1000007#define Maxn 405int c[Maxn+10][Maxn+10];void init(){    memset(c,0,sizeof(c));    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;        }    }}int solve(int m,int n,int k){    if(m*n<k) return 0;    int sum = 0,ans = 0;    for(int s=1;s<16;s++)    {        int row = m;        int col = n;        int t = 0;        if(s&1) row--,t++;        if(s&2) row--,t++;        if(s&4) col--,t++;        if(s&8) col--,t++;        if(t&1) sum = (sum + c[row*col][k])%MOD;        else sum = (sum - c[row*col][k] + MOD)%MOD;    }    ans = (c[m*n][k]%MOD - sum + MOD) % MOD;    return ans;}int main(){    #ifndef ONLINE_JUDGE        freopen("../in.txt","r",stdin);    #endif    int t;    int cas = 0;    int ans;    int m,n,k;    init();    scanf(" %d",&t);    while(t--)    {        cas++;        scanf(" %d %d %d",&m,&n,&k);        ans = solve(m,n,k);        printf("Case %d: %d\n",cas,ans);    }    return 0;}

另外标记一下排列组合C(i,j)的求法，可以通过C(i,j) = C(i-1,j-1) + C(i-1,j)来递推。