Eqs

Eqs

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=15029

题意：

给出系数a1,a2,a3,a4,a5,求出方程a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=0 有多少个解。

样例：

Sample Input

37 29 41 43 47

Sample Output

654

分析：
   如果直接用5个for循环是会超时的，所以把方程a1x1x1x1+a2x2x2x2+a3x3x3x3+a4x4x4x4+a5x5x5x5=0
转变成-（a1x1x1x1+a2x2x2x2）=a3x3x3x3+a4x4x4x4+a5x5x5x5这样就将5层循环减少到3层循环。

 #include<iostream>
 #include<cstring>
 #include<cmath>
 using namespace std;
 const int maxn=;
 int i,a[],count=,sum=;
 short b[maxn];
 int x1,x2,x3,x4,x5;
 int main()
 {
     memset(b,,sizeof(b));
 for(i=;i<;i++)
    cin>>a[i];
 for(x1=-;x1<=;x1++)
 {   if(x1==)  continue;
       for(x2=-;x2<=;x2++)
     {
         if(x2==)  continue;
         sum=-*(a[]*x1*x1*x1+a[]*x2*x2*x2);
         if(sum<)
             b[maxn+sum]++;
         else b[sum]++;
       }
 }
 for(x3=-;x3<=;x3++)
 {
    if(x3==)  continue;
       for(x4=-;x4<=;x4++)
       {
         if(x4==)  continue;
           for(x5=-;x5<=;x5++)
         {
             if(x5==)  continue;
          sum=a[]*x3*x3*x3+a[]*x4*x4*x4+a[]*x5*x5*x5;
              if(sum<)
                 sum=maxn+sum;
               count=count+b[sum];
                 }

     }
     }
 cout<<count<<endl;
 return ;
 }