Eqs(枚举+ hash)

http://poj.org/problem?id=1840

题意：给出系数a1,a2,a3,a4,a5,求满足方程的解有多少组。

思路：有a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0 可得 -（a1x1³+ a2x2³） = a3x3³+ a4x4³+ a5x5³；

先枚举x1,x2,用hash[]记录 sum出现的次数，然后枚举后三个点，若左边出现的sum在右边可以找到，那么hash[sum]即为解的个数。

 #include <cstdio>
 #include <string.h>
 #include <iostream>
 #define N 25000000
 using namespace std;
 short hash[N+];

 int main()
 {
     int a1,a2,a3,a4,a5;
     while(cin>>a1>>a2>>a3>>a4>>a5)
     {
         int x1,x2,x3,x4,x5;
         int ans = ;
         memset(hash,,sizeof(hash));
         for (x1 = -; x1 <= ; x1 ++)
         {
             if(!x1) continue;
             for (x2 = -; x2 <= ; x2 ++)
             {
                 if (!x2) continue;
                 int sum =(a1*x1*x1*x1+a2*x2*x2*x2)*(-);
                 if (sum < )
                     sum += N;
                 hash[sum]++;

             }
         }
         for (x3 = -; x3 <= ; x3 ++)
         {
             if(!x3) continue;
             for (x4 = -; x4 <= ; x4 ++)
             {
                 if (!x4) continue;
                 for (x5 = -; x5 <= ; x5 ++)
                 {
                     if (!x5) continue;
                     int sum = a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5;
                     if (sum < )
                         sum += N;
                     if (hash[sum])
                         ans += hash[sum];
                 }
             }
         }
         cout<<ans<<endl;
     }
     return ;
 }