hdu 1427 速算24点(next_permutation 搜索)

题意：给出四张扑克牌 问能否算出24

思路：http://blog.csdn.net/xingyeyongheng/article/details/11137631

        其实这题只有两种运算顺序

         1（a@b）@c@d

         2 (a@b)@(c@d)

         所以只需要把数字和运算符全排列遍历一边就可以判断出结果

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

int num[5],ok;

const int INF=100000;

int change_num(char a)

{

    if(a=='A') return 1;

    else if(a=='J') return 11;

    else if(a=='Q') return 12;

    else if(a=='K') return 13;

    if(a=='1') return 10;

    else return a-'0';

}

int fun2(int a,int op,int b)

{

    if(op==0)

    {

        return a+b;

    }

    else if(op==1)

    {

        return a-b;

    }

    else if(op==2)

    {

        return a*b;

    }

    else if(op==3)

    {

        if(b==0||a%b!=0 )return INF;

        return a/b;

    }

}

void fun1(int i,int j,int k)

{

    if(ok==1) return;

    int t1,t2,t3;

    t1=fun2(num[0],i,num[1]);

    if(t1!=INF)

    {

        t2=fun2(t1,j,num[2]);

        if(t2!=INF)

        {

            t3=fun2(t2,k,num[3]);



            if(t3==24||t3==-24) ok=1;

        }

    }

    if(ok==0)

    {

        t1=fun2(num[0],i,num[1]);

        t2=fun2(num[2],j,num[3]);

        if(t1!=INF&&t2!=INF)

        {

                t3=fun2(t1,k,t2);

                if(t3==24||t3==-24) ok=1;

        }

    }

}

int main()

{

    int i,j,k;

    char temp[5];

    while(scanf("%s",temp)!=EOF)

    {

        ok=0;

        num[0]=change_num(temp[0]);

        for(i=1;i<=3;i++)

        {

            scanf("%s",temp);

            num[i]=change_num(temp[0]);

        }

        sort(num,num+4);

        do

        {

           for(i=0;i<4;i++)

           {

               for(j=0;j<4;j++)

               {

                   for(k=0;k<4;k++)

                   {

                      fun1(i,j,k);

                   }

               }

           }

           if(ok==1) break;

        }while (next_permutation(num,num+4));

        if(ok==1) printf("Yes\n");

        else printf("No\n");

    }

    return 0;

}