HDOJ 题目1496 Equations（hash）

最新推荐文章于 2018-07-27 11:23:00 发布

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最新推荐文章于 2018-07-27 11:23:00 发布

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本文详细解析了给定形式的二次方程在指定区间内的整数解的数量计算方法，通过实例展示了求解过程，并提供了算法实现。

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Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5860 Accepted Submission(s): 2371

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

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

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ac代码

#include<stdio.h>
#include<string.h>
#include<math.h>
int hash[2000020];
int main()
{
	int a,b,c,d;
	while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
	{
		int i,j;
		memset(hash,0,sizeof(hash));
		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<=100;i++)
			for(j=1;j<=100;j++)
			{
				hash[1000000+i*i*a+j*j*b]++;
			}
		int sum=0;
		for(i=1;i<=100;i++)
			for(j=1;j<=100;j++)
				if(hash[1000000-i*i*c-j*j*d])
					sum+=hash[1000000-i*i*c-j*j*d];
		printf("%d\n",sum*16);
	}
}