9.3.8 Equations

Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 50 Accepted Submission(s): 36

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

思路：Hash

分两部分Hash，完了

最后结果*16 因为每个数正负都可以，2^4=16

/*
Author:wuhuajun
*/
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;

typedef long long ll;
typedef double dd;
const int maxn=2000010;
int ff[maxn];
int *f=ff+1000003;
int a,b,c,d,sum,cnt,ans;

void close()
{
exit(0);
}


void init()
{
    while (scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF)
    {
        if (a>0 && b>0 && c>0 && d>0)
        {
            printf("0
");
            continue;
        }
        if (a<0 && b<0 && c<0 && d<0)
        {
            printf("0
");
            continue;
        }
        memset(ff,0,sizeof(ff));
        for (int x1=1;x1<=100;x1++)
        {
            if (x1==0) continue;
            for (int x2=1;x2<=100;x2++)
            {
                if (x2==0) continue;
                //printf("%d
",a*x1*x1+b*x2*x2);
                f[a*x1*x1+b*x2*x2]++;
            }
        }
        ans=0;
        for (int x3=1;x3<=100;x3++)
        {
            if (x3==0) continue;
            for (int x4=1;x4<=100;x4++)
            {
                if (x4==0) continue;
                sum=c*x3*x3+d*x4*x4;
                //printf("x3:%d x4:%d sum:%d
",x3,x4,sum);
                ans+=f[-sum]*16;
            }
        }
        printf("%d
",ans);
    }
}

int main ()
{
    init();
    close();
    return 0;
}