測试赛C - Eqs（哈希）

C - Eqs

Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Description

Consider equations having the following form: 

a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=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

哈希存下一開始的两个值，看找后面三个的和，看能不能出现0

让5个for循环转化为1个双重for循环+1个三重for循环

#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxint 25000000
using namespace std;
short p[25000001];
int main()
{
    int a, b, c, d, e, i, j, k, n, s;
    while(scanf("%d %d %d %d %d", &a, &b, &c, &d, &e)!=EOF)
    {
        s = 0;
        memset(p,0,sizeof(p));
        for(i = -50; i <= 50 ; i++)
        {
            if(!i)
                continue ;
            for(j=-50; j<=50; j++)
            {
                if(!j)
                    continue ;
                n = i*i*i*a + j*j*j*b ;
                n = -n ;
                if(n<0)
                    n += maxint ;
                p[n]++;
            }
        }
        for(i = -50 ; i <= 50; i++)
        {
            if(!i)
                continue ;
            for(j = -50; j <= 50; j++)
            {
                if(!j)
                    continue ;
                for(k = -50; k <= 50; k++)
                {
                    if(!k)
                        continue ;
                    n= i*i*i*c + j*j*j*d + k*k*k*e;
                    if(n < 0)
                        n += maxint ;
                    if(p[n])
                        s += p[n];
                }
            }
        }
        printf("%d\n",s);
    }
    return 0;
}