Poj 1840 Eqs（Hash）

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 12911		Accepted: 6324

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;

给出一个5元3次方程，输入其5个系数，求它的解的个数

其中系数 ai∈[-50,50]  自变量xi∈[-50,0）∪（0,50]

水题不解释

#include <stdio.h>
int Hash[25000001];

int main()
{
//cout<<Num[11010]<<endl;
int ans=0;
int a1,a2,a3,a4,a5;
while(scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5)==5)
//memset(Hash,0,sizeof(Hash));
for(int x1=-50; x1<=50; x1++){
if(!x1)
continue;
for(int x2=-50; x2<=50; x2++){
if(!x2)
continue;
int pro=(-1)*(a1*x1*x1*x1+a2*x2*x2*x2);
if(pro<0)
pro+=25000000;
Hash[pro]++;
}
}
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 pro=a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5;
if(pro<0)
pro+=25000000;
if(Hash[pro])
ans+=Hash[pro];
}
}
}
printf("%d\n",ans);
return 0;
}