hdu 1496 Equations（暴力，哈希表 剪枝）

Equations

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

Total Submission(s): 5630    Accepted Submission(s): 2237

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

 

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

代码如下：

#include
int main()
{
	int a,b,c,d,i,j,k,count,l,s1,s2;
	int pow[110];
	while(~scanf("%d%d%d%d",&a,&b,&c,&d))
	{
		count=0;
		for(i=1;i<101;++i)
		pow[i]=i*i;
		if(a>0&&b>0&&c>0&&d>0||a<0&&b<0&&c<0&&d<0)
		{
			printf("0\n");
			continue;
		}
		for(i=1;i<101;++i)
		{
			for(j=1;j<101;++j)
			{
				s1=a*pow[i]+b*pow[j];
				if(s1>0&&c>0&&d>0||s1<0&&c<0&&d<0)
				continue;
				for(k=1;k<101;++k)
				{
					s2=s1+c*pow[k];
					if(s2%d==0)
					{
						s2/=-d;
						if(s2>0)
						{
							for(l=1;l<101;++l)
							{
								if(s2==pow[l])
								{
									count++;
									break;
								}
							}
						}
					}
				}
			}
		}
		printf("%d\n",count*16);
	}
	return 0;
}