POJ 1840 Eqs hash

题意：解方程组a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 ,x属于[50,50]且x!=0输入a1,a2,a3,a4,a5，输出一共有多少种满足方程的解。

题解：左右分开，hash

#include <iostream>
using namespace std;

#define prime 14999
int cube[105], hash[prime][100], a[6];

void get_cube()
{
int t = -1;
for( int i = -50; i <= 50; i++ )
{
if ( i == 0 ) continue;
cube[++t] = i*i*i;
}
}

void get_hash()
{
int sum, temp, i, j, k;
for ( i = 0; i < 100; ++i )
{
for ( j = 0; j < 100; ++j )
{
k = 0;
sum = -(cube[i]*a[1] + cube[j]*a[2]);
temp = sum % prime;
if ( temp < 0 ) temp += prime;
while ( hash[temp][k] != -1 ) ++k;
hash[temp][k] = sum;
}
}
}

int eqs ()
{
int i, j, k, h;
int sum, temp, ans = 0;
for ( i = 0; i < 100; ++i )
for ( j = 0; j < 100; ++j )
for ( k = 0; k < 100; ++k )
{
sum = cube[i]*a[3] + cube[j]*a[4] + cube[k]*a[5];
temp = sum % prime;
if ( temp < 0 ) temp += prime;
h = 0;
while ( hash[temp][h] != -1 )
{
if ( hash[temp][h] == sum )
++ans;
++h;
}
}
return ans;
}

int main()
{
memset(cube,0,sizeof(cube));
memset(hash,-1,sizeof(hash));
for ( int i = 1; i <= 5; ++i )
scanf("%d",a+i);
get_cube();
get_hash();
printf("%d\n",eqs());
return 0;
}