HDU 1496

Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3188 Accepted Submission(s): 1247

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

//网上说这样的方法就是哈希，但我感觉像是彻彻底底的二分，因为根本没有hash函数 
//巧妙之处在于区间等都对称 
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int hash1[1000010];//存储大于零的哈希值
int hash2[1000010];//存储小于零的哈希值

int main()
{
     //setbuf(stdout,NULL);//（File *stream,char *buf）设置流使用 buf缓冲区 ,若buf为NULL，则不使用缓冲区 
     int ans,i,j;
     int a,b,c,d;//最大值为20w 
     while(cin>>a>>b>>c>>d)
     {
         if((a>=0&&b>=0&&c>=0&&d>=0)||(a<0&&b<0&&c<0&&d<0))//没有这个会超时 
         {//去他大爷，明明说没有0，不加上0一直wa 
             cout<<0<<endl;
            // printf("0\n");
             continue;
         }
         memset(hash1,0,sizeof(hash1));
         memset(hash2,0,sizeof(hash2));
         for(i=1;i<=100;i++)
         {
             for(j=1;j<=100;j++)
             {
                 int temp=a*i*i+b*j*j;
                 if(temp>=0)//不必加上temp小于10w，因为两个for循环和输入已经限制了 
                 {
                     hash1[temp]++;
                 }
                 else 
                    hash2[-temp]++;
             }
         }
         ans=0;
         for(i=1;i<=100;i++)
         {
             for(j=1;j<=100;j++)
             {
                 int temp=c*i*i+d*j*j;
                 if(temp>0)
                 {
                     ans+=hash2[temp];//别把数组顺序搞反,因为a*x1^2+b*x2^2 = -(c*x3^2+d*x4^2) 
                 }
                 else
                     ans+=hash1[-temp];
             }
         }
         cout<<(ans<<4)<<endl;//每个数都可以取正负，并不是因为解向量排列 
     }
     return 0;
}
//注意：devc++中不能定义全局变量count，它和库函数中的函数名同名了