poj-1840 Eqs

                                                               D - Eqs

Crawling in process...Crawling failedTime Limit:5000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

SubmitStatusPracticePOJ 1840

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

若是按照常規的做法從最小數一個個試的話，結果一定會超限。所以先計算幾個數的值，放在hash表中，在計算另外的數的值，若兩個值相減爲0，就輸出。

#include<iostream>
 using namespace std;
 
 short hash[25000001];
 
 int main(void)
 {
     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;
 }