Eqs poj 1840 哈希

最新推荐文章于 2024-04-12 00:11:17 发布

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本文介绍了一种通过预计算哈希表的方法来解决特定形式的五元立方方程问题，即寻找满足条件的整数解。文章详细展示了如何通过枚举变量并将方程分为两个部分来简化问题，进而有效地计算出所有可能的解。

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total 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+12500000]++;
                }
         }
     }
     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+12500000];
          }
    }
    printf("%d\n",ans);
    return 0;
}