Eqs poj 1840 哈希

Eqs

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 15979 Accepted: 7832

Description

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. 

Input

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

Output

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

Sample Input

37 29 41 43 47

Sample Output

654

#include <iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;char hash[25000005];int a[6];int main(){     int i,j,k;     memset(hash,0,sizeof(hash));     for(i=1;i<6;i++)        scanf("%d",&a[i]);     for(i=-50;i<=50;i++)//我们可以将x1,x2,x3,x4,x5分为两部分，相求x1,x2,x3这部分，     {         if(i==0) continue;         for(j=-50;j<=50;j++)         {                if(j==0)  continue;                for(k=-50;k<51;k++)                {                     if(k==0) continue;                     int t=i*i*i*a[1]+j*j*j*a[2]+k*k*k*a[3];                     if(t>12500000||t<-12500000)                        continue;                     hash[t[phone]]++;                }         }     }     int ans=0;     for(i=-50;i<=50;i++)//枚举x4x5的所有结果;     {          if(i==0)   continue;          for(j=-50;j<=50;j++)          {               if(j==0)  continue;               int t=-(i*i*i*a[4]+j*j*j*a[5]);              ans+=hash[t[phone]];          }    }    printf("%d\n",ans);    return 0;}