HDU 1496 Equations (hash)

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本文探讨了形式为a*x1^2 + b*x2^2 + c*x3^2 + d*x4^2 = 0的四元二次方程的求解方法，其中系数a, b, c, d为[-50,50]区间内的非零整数。通过枚举和哈希技术确定方程的有效整数解数量。

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Equations

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

Total Submission(s): 92 Accepted Submission(s): 58

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

Recommend

暴力着hash 4重循环肯定TL，可以把表达式通分一下， a*x1^2+b*x2^2 = -c*x3^2-d*x4^2 然后hash见代码。

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;
int sq[105];
constexpr int maxn = 1000005;
int h[2000010];
int main() {
    for (int i = 1; i <= 100; ++i) {
        sq[i] = i*i;
    }
    int a, b, c, d;
    while (cin >> a >> b >> c >> d) {
        memset(h, 0, sizeof(h));
        if (a > 0 && b > 0 && c > 0 && d > 0) {
            cout << 0 << endl;
        } else {
            for (int i = 1; i <= 100; ++i) {
                for (int j = 1; j <= 100; ++j) {
                    h[a * sq[i] + b * sq[j] + maxn]++;
                }
            }
            int ans = 0;
            for (int i = 1; i <= 100; ++i) {
                for (int j = 1; j <= 100; ++j) {
                    ans += h[-c * sq[i] - d * sq[j] + maxn];
                }
            }
            cout << ans * 16 << endl;//4个整数的正负 2 ^ 4
        }
    }
}