A 2 3 6 3 3 8 8
Yes No
算法分析：
首先进行全排列，之后进行深搜，注意括号只有两种情况（a@b）@(c@d)和(（a@b）@c)@d.
#include<iostream>
#include<algorithm>
using namespace std;
int res[4];
void dfs(int sum, int cur, int temp);
bool flag = 0;
int main()
{
char s[3];
int i,j,k;
while(1)
{
for(i = 0;i < 4;i ++)
{
if(scanf("%s",s) == EOF) return 0;
if(s[0] == 'A') res[i] = 1;
else if(s[0] == 'J') res[i] = 11;
else if(s[0] == 'Q') res[i] = 12;
else if(s[0] == 'K') res[i] = 13;
else if(s[0] == '1' && s[1] == '0') res[i] = 10;
else res[i] = s[0] - '0';
}
sort(res, res+4);
flag = 0;
do
{
dfs(res[0], 1, res[1]);
}while(next_permutation(res, res+4)&&!flag);
if(flag) printf("Yes\n");
else printf("No\n");
}
return 0;
}
void dfs(int sum, int cur, int temp)
{
if(flag)
return;
if(cur==3)
{
if(sum+temp==24)
flag=1;
if(sum-temp==24)
flag=1;
if(sum*temp==24)
flag=1;
if(temp!=0&&sum%temp==0&&sum/temp==24)
flag=1;
return;
}
dfs(sum+temp, cur+1, res[cur+1]);
dfs(sum-temp, cur+1, res[cur+1]);
dfs(sum*temp, cur+1, res[cur+1]);
if(temp!=0&&sum%temp==0)
dfs(sum/temp, cur+1, res[cur+1]);
dfs(sum, cur+1, temp+res[cur+1]);
dfs(sum, cur+1, temp-res[cur+1]);
dfs(sum, cur+1, temp*res[cur+1]);
if(res[cur+1]!=0&&temp%res[cur+1]==0)
dfs(sum, cur+1, temp/res[cur+1]);
}另一种解法，#include <stdio.h>
#include <string.h>
int a[10];
int map[10];
int flag = 1;
int get(char *str)
{
if(str[0] == 'A')
return 1;
if(str[0] == '1')
return 10;
if(str[0] == 'J')
return 11;
if(str[0] == 'Q')
return 12;
if(str[0] == 'K')
return 13;
return str[0] - '0';
}
void DFS(int s1,int s2,int k,int x)//(A@B)@(C@D)
{
int i;
if(k == 4)
{
if(x == 1)
{
if(s1 == 24 || s1 == -24)
{
flag = 1;
}
}
else
{
if(s1 + s2 == 24 || s1 + s2 == -24 || s1 - s2 == 24 || s1 - s2 == -24 || s1 * s2 == 24 || s1 * s2 == -24)
flag = 1;
if(s2 && s1 % s2 == 0 && s1 /s2 == 24)
{
flag = 1;
}
}
return ;
}
for(i = 1;i <= 4; i++)
{
if(!map[i])
{
map[i] = 1;
if(k == 0)
{
DFS(a[i],s2,k+1,1);
}
else if(k == 1)
{
DFS(s1+a[i],s2,k+1,1);
DFS(s1-a[i],s2,k+1,1);
DFS(s1*a[i],s2,k+1,1);
if(a[i] && s1 % a[i] == 0)
DFS(s1/a[i],s2,k+1,1);
}
else
{
if(k == 2)
{
DFS(s1+a[i],s2,k+1,1);
DFS(s1-a[i],s2,k+1,1);
DFS(s1*a[i],s2,k+1,1);
if(a[i] && s1 % a[i] == 0)
DFS(s1/a[i],s2,k+1,1);
DFS(s1,a[i],k+1,2);
}
else
{
if(x == 1)
{
DFS(s1+a[i],s2,k+1,1);
DFS(s1-a[i],s2,k+1,1);
DFS(s1*a[i],s2,k+1,1);
if(a[i] && s1 % a[i] == 0)
DFS(s1/a[i],s2,k+1,1);
}
else
{
DFS(s1,s2+a[i],k+1,2);
DFS(s1,s2-a[i],k+1,2);
DFS(s1,s2*a[i],k+1,2);
if(a[i] && s2 % a[i] == 0)
DFS(s1,s2 / a[i],k+1,2);
}
}
}
map[i] = 0;
}
}
}
int main()
{
char str[12];
int i;
while(scanf("%s",str)!=EOF)
{
flag = 0;
a[1] = get(str);
for(i = 2;i <= 4;i++)
{
scanf("%s",str);
a[i] = get(str);
}
memset(map,0,sizeof(map));
DFS(0,0,0,1);
if(flag)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}C++：
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int main()
{
string str;
cin >> str;
sort(str.begin(), str.end());
cout << str << endl;
while (next_permutation(str.begin(), str.end()))
{
cout << str << endl;
}
return 0;
}C:
#include <cstdio>
#include <algorithm>
#include <cstring>
#define MAX 100
using namespace std;
int main()
{
int length;
char str[MAX];
gets(str);
length = strlen(str);
sort(str, str + length);
puts(str);
while (next_permutation(str, str + length))
{
puts(str);
}
return 0;
}原文地址：http://blog.csdn.net/u014492609/article/details/40268385
