POJ 1840, Eqs

Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 3409 Accepted: 1514

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

Source
Romania OI 2002

// POJ1840.cpp : Defines the entry point for the console application.
//

#include <iostream>
using namespace std;

struct Node
{
Node():count(0), num(0), next(NULL){}
int num;
int count;
Node* next;
};

int main(int argc, char* argv[])
{
int a[5];
cin >> a[0] >> a[1] >> a[2] >> a[3] >> a[4];

const int SIZE = 33119;
Node hash[SIZE];

//create lookup table
int X[51];
for (int i = 0; i <=50; ++i) X[i] = i * i * i;

//init hash table
for (int i = -50; i <=50; ++i)
for (int j = -50; j <= 50; ++j)
if (i != 0 && j != 0)
{
int num = a[0] * (i < 0 ? -X[-i]:X[i]) + a[1] * (j < 0 ? -X[-j]:X[j]);
int key = num % SIZE;
if (key < 0) key += SIZE;
Node* pt = &hash[key];
bool found = false;
while (pt->next != NULL)
{
pt = pt->next;
if (pt->num == num)
{
++(pt->count);
found = true;
break;
}
}
if (found == false)
{
Node* pc = new Node;
pc->count = 1;
pc->num = num;
pt->next = pc;
};
};

//search hash table and find match
int cnt = 0;
for (int i = -50; i <=50; ++i)
for (int j = -50; j <= 50; ++j)
for (int k = -50; k <= 50; ++k)
if (i != 0 && j != 0 && k != 0)
{
int num = -(a[2] * (i < 0 ? -X[-i]:X[i]) +
　　　　　　　　　　a[3] * (j < 0 ? -X[-j]:X[j]) + a[4] * (k < 0 ? -X[-k]:X[k]));
int key = num % SIZE;
if (key < 0) key += SIZE;

Node* pt = &hash[key];
while (pt->next != NULL)
{
pt = pt->next;
if (pt->num == num)
{
cnt += pt->count;
break;
}
}
};

cout << cnt <<endl;
return 0;
}