HDU1496 Equations 卡时间第二题

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.

InputThe 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.OutputFor 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

#include<cstdio>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#include<memory>

#include<map>

#include<cstring>

using namespace std;

int a,b,c,d;

int q[];

int main()

{

    int i,j,k,ans;

    while(cin>>a>>b>>c>>d){

        ans=;

        if(a>&&b>&&c>&&d>){

          printf("0\n");

          continue;

        }

        if(a<&&b<&&c<&&d<){

          printf("0\n");

          continue;

        }

        memset(q,,sizeof(q));

        for(i=;i<=;i++)

         for(j=;j<=;j++)

           q[i*i*a+j*j*b+]++;

        for(i=;i<=;i++)

         for(j=;j<=;j++)

           ans+=q[-i*i*c-j*j*d];

        printf("%d\n",ans*);

    }

    return ;

}

#include<cstdio>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#include<memory>

#include<map>

using namespace std;

int a,b,c,d;

map<int,int>q;

int main()

{

    int i,j,k,ans;

    while(cin>>a>>b>>c>>d){

        ans=;

         if(a>&&b>&&c>&&d>||a<&&b<&&c<&&d<)  //因为x那项永远是正数，如果系数都为正或者为负的时候明显不等于0

        {

            printf("0\n");

            continue;

        }

        q.clear();

        for(i=;i<=;i++)

         for(j=;j<=;j++)

           q[i*i*a+j*j*b]++;

        for(i=;i<=;i++)

         for(j=;j<=;j++)

           ans+=q[-i*i*c-j*j*d];

        printf("%d\n",ans*);

    }

    return ;

}

巴特西

HDU1496 Equations 卡时间第二题

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