hdu1496——哈希妙用（暴力+优化）

本文链接：https://blog.youkuaiyun.com/qq_43472263/article/details/99420208

本文介绍了一种使用哈希表快速解决四元二次方程问题的方法，通过预计算并存储可能的解组合，实现了高效查找匹配解的目标。文章详细解释了算法流程，并提供了完整的C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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

这个题直接暴力求x1,x2,x3，算出x4,判断一下，再加一个特判就可以过，不过用哈希来实现更简单，

#include <iostream>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;
const int maxn=1e6+7;
int hash1[maxn],hash2[maxn];
int main(int argc, char** argv) {
	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&&d<0&&c<0)){
			printf("0\n");
			continue;
		} 
		int ans=0;
		memset(hash1,0,sizeof(hash1));
		memset(hash2,0,sizeof(hash2));
		for(int i=1;i<=100;++i){
			for(int j=1;j<=100;++j){
				int x=a*i*i+b*j*j;
				if(x>=0) hash1[x]++;
				else hash2[-x]++;
			}
		}
		for(int i=1;i<=100;++i){
			for(int j=1;j<=100;++j){
				int x=c*i*i+d*j*j;
				if(x>0) ans+=hash2[x];
				else ans+=hash1[-x];
			}
		}
		printf("%d\n",ans*16);
	}
	return 0;
}