Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9370 Accepted Submission(s): 3895

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

问题链接：HDU1496 Equations

问题简述：（略）

问题分析：

做两个哈希表来降低组合数。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* HDU1496 Equations */

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

const int N = 100;
const int N2 = N * N * N;
int sum1[N2 + 1], sum2[N2 + 1];

int main()
{
    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 && c < 0 && d < 0)) {
            printf("0\n");
            continue;
        }

        memset(sum1, 0, sizeof(sum1));
        memset(sum2, 0, sizeof(sum2));

        int sum = 0;
        for(int i = 1; i <= N; i++)
            for(int j = 1; j <= N; j++) {
                int k = a * i * i + b * j * j;
                if(k >= 0)
                    sum1[k]++;
                else
                    sum2[-k]++;
            }
        for(int i = 1; i <= N; i++)
            for(int j = 1; j <= N; j++) {
                int k = c * i * i + d * j * j;
                if(k > 0)
                    sum += sum2[k];
                else
                    sum += sum1[-k];
            }

        // 每个解有正有负，所以结果有2^4种
        printf("%d\n", 16 * sum);
    }

    return 0;
}

HDU1496 Equations【哈希表】

Equations

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