POJ 1840 Simple Hash

http://poj.org/problem?id=1840

Description

Consider equations has the following form:
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0
The coefficients is 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 would contain on the first line the number of the solutions for the given equation.

Sample Input

37 29 41) 43 47

Sample Output

654

/**poj 1840 hash*/#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm >using namespace Std;typedef long long ll;const int maxn=25000008;int a1,a2,a3,a4,a5;short hash[maxn*2];///int It's going to explode. The space complexity of the memory constraint is tight. int main () {while (CIN&GT;&GT;A1&GT;&GT;A2&GT;&GT;A3&GT;&GT;A4&GT;&GT;A5) {for (int i=-50;i&l            t;=50;i++) {if (i==0) continue;                for (int j=-50;j<=50;j++) {if (j==0) continue;                    for (int k=-50;k<=50;k++) {if (k==0) continue;                hash[i*i*i*a1+j*j*j*a2+k*k*k*a3+maxn]++;        }}} LL ans=0;            for (int i=-50;i<=50;i++) {if (i==0) continue;                for (int j=-50;j<=50;j++) {if (j==0) continue;            ANS+=HASH[-I*I*I*A4-J*J*J*A5+MAXN];    }} cout<< ans<< Endl; } return 0;}

POJ 1840 Simple Hash