POJ 1840 Eqs hash

本文主要是介绍POJ 1840 Eqs hash，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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;
}

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