HDU 1496 Equations (hash)

最新推荐文章于 2018-06-05 10:19:00 发布

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本文讨论了如何通过枚举法解决特定形式的二次方程组，并计算其整数解的数量。通过引入哈希表来优化计算过程，提高了效率。

Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3625 Accepted Submission(s): 1444

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

/*
  由a*x1^2 + b*x2^2 + c*x3^2 + d*x4^2 = 0移项得到
  a*x1^2 + b*x2^2 = -(c*x3^2 + d*x4^2);考虑到-100<=x<=100,
  则可以枚举式子两边可取的值，最后统计结果。统计则用到
  hash，将能计算出来的结果作为hash的索引。首先枚举左式，若结果为正，
  存到left_hash[]，为负则存到right_hash[]，其中统计的信息是该结
  果出现的次数。枚举右式时，若结果为正，则加上right_hash[]，
  因为right_hash[]存的是左式为负数时的情况，同理，若右式结果为负
  则加上left_hash[]最后结果ans * 16即是答案。因为x1,x2,x3,x4都
  是取平方的，每一个都可以取正或者取负，有2^4种情况。
  
  281MS	8084K
  */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define SIZE 1000005  //最大索引为100*100*50 + 100*100+50

using namespace std;

int left_hash[SIZE];
int right_hash[SIZE];
int a,b,c,d;
int ans;

int main()
{
    while(~scanf("%d%d%d%d",&a,&b,&c,&d))
    {
        if((a >=0 && b>=0 && c>=0 && d>=0) ||
                (a<0 && b<0 && c<0 && d<0))
        {
            printf("0\n");
            continue;
        }
        for(int i=0; i<SIZE; i++)
            left_hash[i] = right_hash[i] = 0;
        ans = 0;
        for(int i=1; i<=100; i++)
        {
            for(int j=1; j<=100; j++)
            {
                int temp = a*i*i + b*j*j;
                if(temp >= 0)
                    left_hash[temp] ++;
                else
                    right_hash[-temp] ++;
            }
        }
        for(int i=1; i<=100; i++)
        {
            for(int j=1; j<=100; j++)
            {
                int temp = c*i*i + d*j*j;
                if(temp > 0)
                    ans += right_hash[temp];
                else
                    ans += left_hash[-temp];
            }
        }
        printf("%d\n",ans*16);
    }
    return 0;
}