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校園文化活動月”之“校慶杯”大學生程序設計競賽暨杭州電子科技大學第四屆大學生程序設計競賽 代碼如下：

#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;
}