Description

考虑方程a1a_1a1​x13 x_1^3x13​ +a2a_2a2​x23 x_2^3x23​ +a3a_3a3​ x33x_3 ^3x33​=000，系数是在[-50,50]区间的整数，xi∈[−50,50]x_i∈[-50,50]xi​∈[−50,50]，xi!=0x_i!=0xi​!=0，i∈1,2,3,4,5i∈{1,2,3,4,5}i∈1,2,3,4,5，求满足方程的解的数量。

Format

Input

唯一的输入行包含5个系数a1、a2、a3、a4、a5a_1 、a_2 、a_3 、a_4 、a_5a1​、a2​、a3​、a4​、a5​ ，以空格分隔。.

Output

单行输出满足方程的解的数量。

Samples

##输入数据 1

37 29 41 43 47

输出数据 1

654

#include <iostream>
#include <cstring>
using namespace std;
#define N 31250000
short hashs[N];
short hashsfu[N];
int main(){
    int a1,a2,a3,a4,a5;
    long the_end;
    cin >> a1 >> a2 >> a3 >> a4 >> a5;
    memset(hashs,0,sizeof(hashs));
    for (int i = -50; i <= 50; i++)
    {
       for (int j = -50; j <= 50; j++)
       {
         for (int z = -50; z <= 50; z++)
         {
            if(i == 0 || j==0 || z == 0) continue;
            the_end = i*i*i*a1 + j*j*j *a2 + z*z*z*
            a3;
            if(the_end < 0 ){
                hashsfu[the_end*-1]++;
            }else{
                hashs[the_end]++;
            }
         }
         
       }
       
    }

    long ans = 0;
    for (int i = -50; i <= 50; i++){
        for (int j = -50; j <= 50; j++){
            if(i==0 || j == 0) continue;
             the_end = -1*(i*i*i*a4 + j*j*j*a5);
            if(the_end < 0 ){
                ans+=hashsfu[the_end*-1];
            }else{
                ans+=hashs[the_end];
            }
             
        }
    }
    cout << ans ;

    
}