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  • (hash) hdu 1496

    Equations

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 6089    Accepted Submission(s): 2465


    Problem Description
    Consider equations having the following form: 

    a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
    a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.

    It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.

    Determine how many solutions satisfy the given equation.
     
    Input
    The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks.
    End of file.
     
    Output
    For each test case, output a single line containing the number of the solutions.
     
    Sample Input
    1 2 3 -4 1 1 1 1
     
    Sample Output
    39088 0
     
    Author
    LL
     
    Source
     
     
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<string>
    #include<algorithm>
    #include<cstdlib>
    using namespace std;
    int hash1[1000005],hash2[5000005];
    int a,b,c,d;
    int main()
    {
        while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
        {
            int ans=0,temp;
            if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
            {
                printf("0
    ");
                continue;
            }
            for(int i=0;i<1000005;i++)
                hash1[i]=hash2[i]=0;
            for(int i=1;i<=100;i++)
            {
                for(int j=1;j<=100;j++)
                {
                    temp=a*i*i+b*j*j;
                    //printf("%d
    ",temp);
                    if(temp>=0)
                        hash1[temp]++;
                    else
                        hash2[-temp]++;
                }
            }
            for(int i=1;i<=100;i++)
            {
                for(int j=1;j<=100;j++)
                {
                    temp=c*i*i+d*j*j;
                    if(temp>0)
                        ans+=hash2[temp];
                    else
                        ans+=hash1[-temp];
                }
                //printf("%d
    ",ans);
            }
            printf("%d
    ",16*ans);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/water-full/p/4473478.html
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