POJ 1840 --EQS

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Eqs

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 13934 Accepted: 6840

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

首先 题目的意思就是解一个 五元3次的方程 求满足等式的解的个数；

直接暴力循环求解呢，肯定会超时的；

可以将5层循环 拆分成一个2层的和一个3层的 他们的值是相反数 ，这样的话时间会减少；

保存前2项和的同时 也要记得保存 出现这个和的可能的解得个数；

也就是 hash 求值 ；

当然，注意一个特殊条件 ， 各值不能存在0；

#include <iostream>#include <cstring>using namespace std;short hash[25000001]; //hash [sum] 表示值等于sum 的解的数目int main(){    int a1, a2, a3, a4, a5;    while ( cin >> a1>>a2>>a3>>a4>>a5){        memset(hash, 0, sizeof(hash)) ;        for ( int x1 = -50 ; x1 <= 50; x1 ++ )        {            if ( !x1)                continue ;            for ( int x2 = -50 ; x2 <= 50 ; x2 ++)            {                if (!x2)                    continue;                int sum = (a1 *x1*x1*x1 + a2*x2*x2*x2) *(-1);                if ( sum < 0 )                    sum += 25000000;                hash[sum] ++ ;            }        }        int solution = 0 ;        for ( int x3 = -50 ; x3 <=50 ; x3 ++ )        {            if (!x3)                continue ;            for ( int x4 = -50 ; x4 <= 50 ; x4 ++ )            {                if (!x4)                    continue ;                for ( int x5 = -50 ; x5 <= 50 ; x5 ++ )                {                    if (!x5)                        continue ;                    int sum = a3*x3*x3*x3 + a4 *x4*x4*x4 + a5*x5*x5*x5 ;                    if ( sum < 0 )                        sum += 25000000;                    if ( hash[sum ])                        solution += hash[sum ];                }            }        }        cout <<solution << endl ;    }    return 0;}