Poj 1840 simple hash and poj1840hash

Poj 1840 simple hash and poj1840hash

Http://poj.org/problem? Id = 1840

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, xiε [-50, 50], xi! = 0, any I ε {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 in 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 LL; const int maxn = 25000008; int a1, a2, a3, a4, a5; short hash [maxn * 2]; // int will kill the memory constraint. The space complexity is very tight. int main () {while (cin> a1> a2> a3> a4> a5) {for (int I =-50; I <= 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 * a1 + j * a2 + 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 * a4-j * j * a5 + maxn];} cout <ans <endl;} return 0 ;}