(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

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

 

 

#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\n");
            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\n",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\n",ans);
        }
        printf("%d\n",16*ans);
    }
    return 0;
}