POJ 1840 Eqs 暴力

最新推荐文章于 2021-09-30 13:50:21 发布

转载最新推荐文章于 2021-09-30 13:50:21 发布 · 56 阅读

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原文链接：http://www.cnblogs.com/Noevon/p/5683610.html

本文介绍了一种解决特定形式五次方程的方法，通过预处理和查询技术减少计算复杂度，实现高效求解。

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

Source

Romania OI 2002

题目大意：给a1, a2, a3, a4, a5,5个数且 xi∈[-50,50],xi != 0, any i∈{1,2,3,4,5}.求出有多少种解使得a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0

题解：5个for肯定会超时(1e8),可以分开考虑，移项可以得到-a1x1³- a2x2³= a3x3³+ a4x4³+ a5x5³,把左边所有的可能存在数组里，右边查询即可,左边可能为负数，所以两边同时加上25001000

#include <stdio.h>
char mp[70000000];
int main()
{
    int t,i, j, k,a2, a1, a3,a4,a5;
    scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
        for(j=-50;j<=50;j++)
            for(k=-50;k<=50;k++)
                if(j!=0&&k!=0)
                    mp[a1*j*j*j+a2*k*k*k+25001000]++;
    for(t=0,i=-50;i<=50;i++)
        for(j=-50;j<=50;j++)
            for(k=-50;k<=50;k++)
                if(i!=0&&j!=0&&k!=0)
                    t += mp[25001000-a3*k*k*k-i*i*i*a4-j*j*j*a5];
    printf("%d\n", t);
return 0;
}