Eqs

最新推荐文章于 2024-04-12 00:11:17 发布

lucky_vikey

最新推荐文章于 2024-04-12 00:11:17 发布

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Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 12043		Accepted: 5891

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

根据题意将一个五次循环分解成一个三次循环和二次循环

<span style="font-size:18px;color:#009900;">#include <iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
short f[25000001];
#define Max 25000000
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(f,0,sizeof(f));
        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 += Max ;
                f[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 += Max ;
                    if(f[n])
                        s += f[n];
                }
            }
        }
        printf("%d\n",s);
    }
    return 0;
}
</span>