Equations（hdu 1496 二分查找+各种剪枝）

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6927    Accepted Submission(s): 2810

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

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
int a,b,c,d;
int s[101];
int search(int mid,int k)
{
    return c*s[k]+d*s[mid];
}
void init()
{
    for(int i=1;i<=100;i++)    s[i]=i*i;
}
int main()
{
    int i,j,k,ans,y;
    freopen("in.txt","r",stdin);
    init();
    while(cin>>a>>b>>c>>d)
    {
        ans=0;
        if(a>0&&b>0&&c>0&&d>0||a<0&&b<0&&c<0&&d<0){
            printf("0
");
            continue;
        }
        for(i=1;i<=100;i++)
        {
            for(j=1;j<=100;j++)
            {
                int sum=a*s[i]+b*s[j];
                if(sum>0&&c>0&&d>0||sum<0&&c<0&&d<0)    continue;
                for(k=1;k<=100;k++)
                {
                    int p=0-sum-c*s[k];
                    if(p%d==0)
                    {
                        p=p/d;
                        if(binary_search(s+1,s+101,p))    ans++;
                    }
                }
            }
        }
        cout<<ans*16<<endl;
    }
}