Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 15010		Accepted: 7366

Consider equations having the following form: 
a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0 
The coefficients are given integers from the interval [-50,50]. 
It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}.

Determine how many solutions satisfy the given equation. 

The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

The output will contain on the first line the number of the solutions for the given equation.

37 29 41 43 47

654

 /*

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #include<list>

 using namespace std;

 const int mxn=14997;

 list<int>ha[mxn*2];

 list<int>::iterator it;

 int x,x1,x2,x3,x4,x5;

 int ans=0;

 int main(){

     scanf("%d%d%d%d%d",&x1,&x2,&x3,&x4,&x5);

     int i,j,k;

     for(i=-50;i<=50;i++)

       for(j=-50;j<=50;j++)

           for(k=-50;k<=50;k++){

               if(i==0||j==0||k==0)continue;

               x=i*i*i*x1+j*j*j*x2+k*k*k*x3;

               ha[x%mxn+mxn].push_back(x);//hash //x%mxn+mxn保证hash完以后是正数

           }

     for(i=-50;i<=50;i++)

       for(j=-50;j<=50;j++){

           if(i==0|j==0)continue;

           x=-(i*i*i*x4+j*j*j*x5);

           //检查hash

         for(it=ha[x%mxn+mxn].begin();it!=ha[x%mxn+mxn].end();it++ ){

             if(*it==x)ans++;

         }

       }

     printf("%d",ans);

     return 0;

 }

 */

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #include<list>

 using namespace std;

 const int mxn=;

 list<int>ha[mxn*];

 list<int>::iterator it;

 int x,x1,x2,x3,x4,x5;

 int ans=;

 int main(){

     scanf("%d%d%d%d%d",&x1,&x2,&x3,&x4,&x5);

     int i,j,k;

     for(i=-;i<=;i++)

       for(j=-;j<=;j++)

           {

               if(i==||j==)continue;

               x=i*i*i*x1+j*j*j*x2;

               ha[x%mxn+mxn].push_back(x);//hash //x%mxn+mxn保证hash完以后是正数

           }

     for(i=-;i<=;i++)

       for(j=-;j<=;j++)

           for(k=-;k<=;k++){

               if(i==|j==||k==)continue;

               x=-(i*i*i*x3+j*j*j*x4+k*k*k*x5);

               //检查hash

             for(it=ha[x%mxn+mxn].begin();it!=ha[x%mxn+mxn].end();it++ ){

                 if(*it==x)ans++;

             }

           }

     printf("%d",ans);

     return ;

 }