HDU 1496 Equations

最新推荐文章于 2018-08-11 23:39:48 发布

窦小雨

最新推荐文章于 2018-08-11 23:39:48 发布

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分类专栏： Hash

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Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 77 Accepted Submission(s): 51

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

题目大意：

给出方程的 abcd 求方程有几个解。

思路：

直接暴力显然是不行的 100 * 100 *100 * 100 直接炸。

hash 一下就可以了，注意 x 1 = 3 x2 = 4 和 x1 =4 x2=3 是不一样的，一开始我的思路是左边两个hash 一下，右边直接 find，但是这样会少好多结果。所以还是用最基本的想法。

最后 * 16 是因为只考虑了一半负数和正数是一样的，

AC代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <memory.h>
using namespace std;

int f1[1000005]; 
int f2[1000005];   
int main()
{
    int i, j, k, sum;
    int a, b, c, d;
    while(scanf("%d %d %d %d", &a, &b, &c, &d) != EOF)
    {
        if(a>0 && b>0 && c>0 && d>0 || a<0 && b<0 && c<0 && d<0)
        {
            printf("0\n");
            continue;
        }
        memset(f1, 0, sizeof(f1));
        memset(f2, 0, sizeof(f2));
        for(i = 1; i <= 100; i++)
        {
            for(j = 1; j<= 100; j++)
            {
                k = a*i*i + b*j*j;
                if(k >= 0) f1[k]++; 
                else f2[-k]++;   
            }
        }
        sum = 0;
        for(i = 1; i <= 100; i++)
        {
            for(j = 1; j<= 100; j++)
            {
                k = c*i*i + d*j*j;
                if(k > 0) sum += f2[k]; 
                else sum += f1[-k];
            }
        }
        printf("%d\n", 16*sum);
    }

    return 0;
}