pku grid 2787 24点问题

#include <iostream>
#include <cmath>
using namespace std;
double number[4];
/*
把这四个数看做一个集合 Ex: A = {1, 2, 3, 4};
取出任意两个数,例如:1+2 = 3, 1-2 = -1, 2-1 = 1, 1/2 = 0.5,  2/1 = 2, 1*2 = 2;
从而得到: A=		  {3,3,4},  {-1,3,4}, {1,3,4}, {0.5,3,4}, {2,3,4}, {2,3,4};
.
.
.
从而不断划归为子问题
*/
bool solve(int left)
{
if(left == 1)
{
if(fabs(number[0] - 24.0) < 1E-6)
return true;
return false;
}
for(int i = 0; i < left; ++i)
for(int j = i+1; j < left; ++j)
{
double a = number[i];
double b = number[j];

number[i] = a + b;
number[j] = number[left-1];
if(solve(left-1))
return true;

number[i] = a - b;
number[j] = number[left-1];
if(solve(left-1))
return true;

number[i] = b- a;
number[j] = number[left-1];
if(solve(left-1))
return true;

number[i] = a * b;
number[j] = number[left-1];
if(solve(left-1))
return true;

if(b)
{
number[i] = a / b;
number[j] = number[left-1];
if(solve(left-1))
return true;
}

if(a)
{
number[i] = b / a;
number[j] = number[left-1];
if(solve(left-1))
return true;
}
number[i] = a;
number[j] = b;
}
return false;
}
int main()
{
while(scanf("%lf%lf%lf%lf", &number[0], &number[1], &number[2], &number[3])
&& !(number[0] == 0 && number[1] == 0 && number[2] == 0 && number[3] == 0))
{
if(solve(4))
printf("YES/n");
else
printf("NO/n");
}
return 0;
}