POJ 1840Eqs（hash）

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 15934		Accepted: 7813

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：给定a1, a2, a3, a4, a5，求a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0的解，x1-x5的范围为[50)U(50]；
题解：用哈兮做，将其分解为a1x13+ a2x23= -（a3x33+ a4x43+ a5x53），定义hash有short；

#include <stdio.h>
#include <string.h>

short hash[25000001];

int main()
{
int a, b, c, d, e, i, j, sum, k, S;
while(~scanf("%d%d%d%d%d",&a, &b, &c, &d, &e))
{
S = 0;
memset(hash, 0, sizeof(hash));
sum = 0;
for(i = -50; i <= 50; i++)
{
if( !i)
continue;
for(j = -50; j <= 50; j++)
{
if(!j)
continue;
sum = a *( i * i * i) + b * ( j * j * j) ;
sum = -sum;
if(sum < 0)
sum += 25000000;
hash[sum]++;
}
}
sum = 0;
for(i = -50; i <= 50; i++)
{
if(!i )
continue;
for(j = -50; j <= 50; j++)
{
if(!j)
continue;
for(k = -50; k <= 50; k++)
{
if(!k )
continue;
sum = c *( i * i * i) + d*( j * j * j) + e* ( k * k * k);
if(sum < 0)
sum += 25000000;
S += hash[sum];
}
}
}
printf("%d\n", S);
}
return 0;
}