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
 1 //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 
 2 //巧妙之处在于区间等都对称 
 3 #include <iostream>
 4 #include <cstdio>
 5 #include <cstring>
 6 using namespace std;
 7 
 8 int hash1[1000010];//存储大于零的哈希值
 9 int hash2[1000010];//存储小于零的哈希值
10 
11 int main()
12 {
13      //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 
14      int ans,i,j;
15      int a,b,c,d;//最大值为20w 
16      while(cin>>a>>b>>c>>d)
17      {
18          if((a>=0&&b>=0&&c>=0&&d>=0)||(a<0&&b<0&&c<0&&d<0))//没有这个会超时 
19          {//去他大爷，明明说没有0，不加上0一直wa 
20              cout<<0<<endl;
21             // printf("0\n");
22              continue;
23          }
24          memset(hash1,0,sizeof(hash1));
25          memset(hash2,0,sizeof(hash2));
26          for(i=1;i<=100;i++)
27          {
28              for(j=1;j<=100;j++)
29              {
30                  int temp=a*i*i+b*j*j;
31                  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 
32                  {
33                      hash1[temp]++;
34                  }
35                  else 
36                     hash2[-temp]++;
37              }
38          }
39          ans=0;
40          for(i=1;i<=100;i++)
41          {
42              for(j=1;j<=100;j++)
43              {
44                  int temp=c*i*i+d*j*j;
45                  if(temp>0)
46                  {
47                      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) 
48                  }
49                  else
50                      ans+=hash1[-temp];
51              }
52          }
53          cout<<(ans<<4)<<endl;//每个数都可以取正负，并不是因为解向量排列 
54      }
55      return 0;
56 }
57 //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

 

 
posted @ 2012-09-02 17:42  加拿大小哥哥  阅读(409)  评论(0)  编辑  收藏  举报