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  • poj 1840 Eqs (hash)

    Eqs

    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

    #include <cstdio>
    #include <cstdlib>
    #include <iostream>
    #include <cstring>
    #include <string>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    const int MAXN=25000010;
    
    short hash[MAXN];
    int a1,a2,a3,a4,a5;
    int ans,temp;
    
    int main()
    {
        while(scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5)!=EOF)
        {
            ans=0;
            memset(hash,0,sizeof(hash));
            for(int i=-50;i<=50;i++)
                for(int j=-50;j<=50;j++)
                {
                    if(i==0||j==0) continue;
                    temp=(a1*i*i*i+a2*j*j*j)*(-1);
                    if(temp<0) temp+=MAXN;
                    hash[temp]++;
                }
            for(int i=-50;i<=50;i++)
                for(int j=-50;j<=50;j++)
                    for(int k=-50;k<=50;k++)
                    {
                        if(i==0||j==0||k==0) continue;
                        temp=a3*i*i*i+a4*j*j*j+a5*k*k*k;
                        if(temp<0) temp+=MAXN;
                        if(hash[temp]) ans+=hash[temp];
                    }
            printf("%d
    ",ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wanghaixv/p/9032572.html
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