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如果三个正整数A B C ,A²+B²=C²则为勾股数
如果ABC之间两两互质，即A与B A与C B与C均互质没有公约数， 则称其为勾股数元组。
请求出给定n m 范围内所有的勾股数元组
输入描述
起始范围 1<n<10000 n<m<10000
输出目描述
abc 保证a<b<c输出格式 a b c
多组勾股数元组 按照a升序b升序 c升序的排序方式输出。
给定范围内，找不到勾股数元组时，输出 Na
案例
输入
1
20
输出
3 4 5
5 12 13
8 15 17
输入
5
10
输出
Na
#include <iostream> #include <vector> #include <math.h> using namespace std;
int gcd(int a, int b) //欧几里得算法 { if (a % b == 0) return b; else return gcd(b, a
% b); } int main() { int n, m; cin >> n >> m; int flag = 0; //标志，判断是否存在，不存在输出Na
for (int i = n; i < m; i++) { for (int j = n + 1; j < m; j++) { for (int k = n +
2; k < m; k++) { if (i * i + j * j == k * k) //勾股数 { if (gcd(i, j) == 1 && gcd(i
, k) == 1 && gcd(j, k) == 1) // 判断是否互质 { flag = 1; if (i < j && j < k) //确定升序 {
cout<< i << " " << j << " " << k << " " << endl; } } } } } } if (flag == 0) cout
<< "Na" << endl; }
