POJ 1840 Eqs

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 8484		Accepted: 4187

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

Source

Romania OI 2002

// 题意很明显求解的个数

// hash保存一半结果再去检测另一半

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define Y 18750003
using namespace std;
char hz[Y],hf[Y];
int main()
{
    int a1,a2,a3,a4,a5;
    int i,j,k,n,t;
    while(scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5)!=EOF)
    {
        for(i=0;i<Y;i++)
          hz[i]=hf[i]=0;
        for(i=-50;i<=50;i++)
          if(i!=0)
          {
              for(j=-50;j<=50;j++)
                if(j!=0)
                 {
                     for(k=-50;k<=50;k++)
                       if(k!=0)
                        {
                            t=a1*i*i*i+a2*j*j*j+a3*k*k*k;
                            if(t>=0)
                              hz[t]++;
                            else
                              hf[-t]++;
                        }
                 }
          }
          n=0;
        for(i=-50;i<=50;i++)
          if(i!=0)
          {
              for(j=-50;j<=50;j++)
                if(j!=0)
                 {
                     t=a4*i*i*i+a5*j*j*j;
                     if(t>0)
                      n+=hf[t];
                     else
                      n+=hz[-t];
                 }
          }
        printf("%d\n",n);

    }
    return 0;
}