五星填数。

 在五星图案节点填上数字：1~12， 不包括7和11。

要求每条直线上数字和相等。

 请搜索所有可能的填法有多少种。

 注意：旋转或镜像后相同的算同一种填法

如下图就是一个恰当的填法。

一、使用dfs算法

#include<iostream>
using namespace std;
const int N=10;
int star[N], num[N]={1, 2, 3, 4, 5, 6, 8, 9, 10, 12};
bool st[N];
int cnt=0, starcnt=0;#define A star[0]+star[2]+star[5]+star[8]
#define B star[1]+star[2]+star[3]+star[4]
#define C star[0]+star[3]+star[7]+star[9]
#define D star[1]+star[5]+star[6]+star[9]
#define E star[4]+star[6]+star[7]+star[8]void printstar()
{cout << "     " << star[0] << "\n";cout << star[1] << "  " << star[2] << "   "<< star[3] << "  " << star[4] << "\n";cout << "  " << star[5] << "     " << star[7] << "\n";cout << "     " << star[6] << "\n";cout << " " << star[8] << "     " << star[9] << "\n\n\n";return;
}void dfs(int u)
{if(u==N){cnt++;if((A==B) && (A==C) && (A==D) && (A==E)){printstar();starcnt++;}return;}for(int i=0; i<N; i++){if(!st[i]){st[i]=1;star[u]=num[i];dfs(u+1);st[i]=0;}}
}int main()
{dfs(0);cout << "total arrange =" << cnt << "\n";cout << " star num=" << starcnt/10 << "\n";return 0;
}

二、使用next_permutation()函数 

#include<iostream>
#include<algorithm>
using namespace std;
const int N=10;
int star[N]={1, 2, 3, 4, 5, 6, 8, 9, 10, 12};
int cnt=0, starcnt=0;#define A star[0]+star[2]+star[5]+star[8]
#define B star[1]+star[2]+star[3]+star[4]
#define C star[0]+star[3]+star[7]+star[9]
#define D star[1]+star[5]+star[6]+star[9]
#define E star[4]+star[6]+star[7]+star[8]void printstar()
{cout << "     " << star[0] << "\n";cout << star[1] << "  " << star[2] << "   "<< star[3] << "  " << star[4] << "\n";cout << "  " << star[5] << "     " << star[7] << "\n";cout << "     " << star[6] << "\n";cout << " " << star[8] << "     " << star[9] << "\n\n\n";return;
}int main()
{do{cnt++;if(A==B && A==C && A==D && A==E){printstar();starcnt++;}}while(next_permutation(star, star+ N));cout << "total arrange =" << cnt << "\n";cout << " star num=" << starcnt/10 << "\n";return 0;
}