Eqs - POJ 1840 哈希

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 11693		Accepted: 5722

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

题意：问有多少组x1-x5使得a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0。

思路：分成前后两部分，注意数据范围，50的4次方*3<2*10的7次方。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
short hash[40000000];
int main()
{ int a1,a2,a3,a4,a5,x1,x2,x3,x4,x5,ans=0;
  scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5);
  for(x1=-50;x1<=50;x1++)
   if(x1!=0)
   for(x2=-50;x2<=50;x2++)
    if(x2!=0)
     hash[(a1*x1*x1*x1+a2*x2*x2*x2)*(-1)+20000000]++;
  for(x3=-50;x3<=50;x3++)
   if(x3!=0)
   for(x4=-50;x4<=50;x4++)
    if(x4!=0)
     for(x5=-50;x5<=50;x5++)
      if(x5!=0)
      ans+=hash[a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5+20000000];
  printf("%d\n",ans);

}