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+1];

int main()
{
    int a1,a2,a3,a4,a5;
    while(cin>>a1>>a2>>a3>>a4>>a5)
    {
        int x1,x2,x3,x4,x5;
        int ans = 0;
        memset(hash,0,sizeof(hash));
        for (x1 = -50; x1 <= 50; x1 ++)
        {
            if(!x1) continue;
            for (x2 = -50; x2 <= 50; x2 ++)
            {
                if (!x2) continue;
                int sum =(a1*x1*x1*x1+a2*x2*x2*x2)*(-1);
                if (sum < 0)
                    sum += N;
                hash[sum]++;

            }
        }
        for (x3 = -50; x3 <= 50; x3 ++)
        {
            if(!x3) continue;
            for (x4 = -50; x4 <= 50; x4 ++)
            {
                if (!x4) continue;
                for (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 += N;
                    if (hash[sum])
                        ans += hash[sum];
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}