http://acm.hdu.edu.cn/showproblem.php?pid=1496

思路：主要将等式化为左右两部分，用一个hash数组先把左边的值存起来，，然后在计算右面的值时只需要寻址就行了，，，，

Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2456 Accepted Submission(s): 969

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

AC代码：

#include<iostream>
#include<string.h>
#define N 1000000
using namespace std;
char Hash[2*N+100];
int main()
{ 
	int a,b,c,d;
	while(~scanf("%d%d%d%d",&a,&b,&c,&d))
	{
		if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))//剪枝，，，，，，，
		{
			printf("0\n");
			continue;
		}
		memset(Hash,0,sizeof(Hash));
		for(int i=1;i<=100;++i)
			for(int j=1;j<=100;++j)
				Hash[1000000+a*i*i+b*j*j]++;
	     int sum=0;
		for(int i=1;i<=100;++i)
			for(int j=1;j<=100;++j)
				sum+=Hash[1000000-c*i*i-d*j*j];
		printf("%d\n",16*sum);
	}return 0;
}