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  • POJ 1840:Eqs 哈希求解五元方程

    Eqs
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 14169   Accepted: 6972

    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

    654

    给定方程是a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 ,给了a1,a2,a3,a4,a5。求解 解的数量。x都在-50到50之间。

    暴搜肯定GG,就得把方程变形将a1x13+ a2x23  移到方程左边,作为结果,然后暴搜x3,x4,x5。时间复杂度就降下来了。 

    代码:

    #include <iostream>
    #include <algorithm>
    #include <cmath>
    #include <vector>
    #include <string>
    #include <cstring>
    #pragma warning(disable:4996)
    using namespace std;
    
    short hash1[25000000];
    
    int main()
    {	
    	int a1, a2, a3, a4, a5;
    	int x1, x2, x3, x4, x5;
    	int temp;
    	long long sum;
    
    	scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5);
    
    	memset(hash1, 0, sizeof(hash1));
    	for (x1 = -50; x1 <= 50; x1++)
    	{
    		if (x1 == 0)continue;
    		for (x2 = -50; x2 <= 50; x2++)
    		{
    			if (x2 == 0)continue;
    			
    			temp = -1*(a1*x1*x1*x1 + a2*x2*x2*x2);
    
    			if (temp < 0)
    			{
    				temp += 25000000;
    			}
    			hash1[temp]++;
    		}
    	}
    	sum = 0;
    	for (x3 = -50; x3 <= 50; x3++)
    	{
    		if (x3 == 0)continue;
    		for (x4 = -50; x4 <= 50; x4++)
    		{
    			if (x4 == 0)continue;
    			for (x5 = -50; x5 <= 50; x5++)
    			{
    				if (x5 == 0)continue;
    				
    				temp = a3*x3*x3*x3 + a4*x4*x4*x4 + a5*x5*x5*x5;
    				
    				if (temp < 0)
    				{
    					temp += 25000000;
    				}
    				if (temp >= 0 && temp < 25000000) 
    				{
    					sum += hash1[temp];
    				}
    			}
    		}
    	}
    	cout << sum << endl;
    	return 0;
    }
    


    版权声明:本文为博主原创文章,未经博主允许不得转载。

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  • 原文地址:https://www.cnblogs.com/lightspeedsmallson/p/4899591.html
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