poj1840 Eqs

使用hash实现，原式：a1*x1^3+a2*x2^3+a3*x3^3+a4*x4^3+a5*x5^3=0

可以推出：a1*x1^3+a2*x2^3 = -(a3*x3^3+a4*x4^3+a5*x5^3) 把O(n^5)转化为O(n^3)+O(n^2)了

#include<iostream>
using namespace std;
short h[25000001];
int main()
{
int a1,a2,a3,a4,a5;
while(cin>>a1>>a2>>a3>>a4>>a5)
{
memset(h,0,sizeof(h));

for(int x1=-50;x1<=50;x1++)
{
if(!x1)
continue;

for(int x2=-50;x2<=50;x2++)
{
if(!x2)
continue;
int sum=(a1*x1*x1*x1 + a2*x2*x2*x2)*(-1);
if(sum<0)
sum+=25000000;
h[sum]++;
}
}

int ans=0;

for(int x3=-50;x3<=50;x3++)
{
if(!x3)
continue;
for(int x4=-50;x4<=50;x4++)
{
if(!x4)
continue;
for(int x5=-50;x5<=50;x5++)
{
if(!x5)
continue;
int sum=a3*x3*x3*x3 + a4*x4*x4*x4 + a5*x5*x5*x5;
if(sum<0)
sum+=25000000;
if(h[sum])
ans+=h[sum];
}
}
}
cout<<ans<<endl;
}
return 0;
}