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hdu 1496 hash+暴力

2024-08-10 11:59:43   240人阅读

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

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
这道题可以用枚举二分来，但是哈希更加简单

我们把每个数都加上1000000就能保证a*i*i, b*j*j, -c*i*i,-d*j*j是正整数了，因此就可以用数组来表示了

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
int hash[1000000*2+7];
int a,b,c,d;
int main()
{
    while(~scanf("%d%d%d%d",&a,&b,&c,&d))
    {
        memset(hash,0,sizeof(hash));
        for(int i=1;i<=100;i++)
        {
            for(int j=1;j<=100;j++)
            {
                hash[i*i*a+j*j*b+1000000]++;
            }
        }
        int sum=0;
        for(int i=1;i<=100;i++)
        {
            for(int j=1;j<=100;j++)
            {
                sum+=hash[-i*i*c-j*j*d+1000000];
            }
        }
        printf("%d\n",sum*16);
    }
    return 0;
}


hdu 1496 hash+暴力


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