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  • HDU 1496 Equations (HASH)

    Equations

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3853    Accepted Submission(s): 1551


    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
     
    Recommend
    LL
     
     
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    
    using namespace std;
    
    const int N=2000010;
    
    int hash[N];
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        int a,b,c,d;
        while(~scanf("%d%d%d%d",&a,&b,&c,&d)){
            if((a>0 && b>0 && c>0 && d>0) || (a<0 && b<0 && c<0 && d<0)){
                printf("0\n");
                continue;
            }
            memset(hash,0,sizeof(hash));
            for(int i=1;i<=100;i++)
                for(int j=1;j<=100;j++)
                    hash[a*i*i+b*j*j+1000000]++;
            int ans=0;
            for(int i=1;i<=100;i++)
                for(int j=1;j<=100;j++)
                    ans+=hash[1000000-(c*i*i+d*j*j)];
            printf("%d\n",ans*16);   //每个解有正有负,结果有2^4种 
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jackge/p/3092983.html
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