Codeforces 707C. Pythagorean Triples

Posted 2020-08-05 Ashly

题目链接:http://codeforces.com/problemset/problem/707/C

题意:

给你直角三角形其中一条边的长度,让你输出另外两条边的长度.

思路:

直接构造勾股数即可,构造勾股数的方法:

　　当 a 为大于 1 的奇数 2 * n+1 时, b = 2 * n * n + 2 * n, c = 2 * n * n + 2 * n + 1.

　　当 a 为大于 4 的偶数 2 * n 时，b = n * n - 1, c = n * n + 1.

然后不满足上述构造方法的数 1, 2, 4直接特判就好.

代码：

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int MAXN = 100000;
 6 typedef long long LL;
 7 
 8 int main() {
 9     ios_base::sync_with_stdio(0); cin.tie(0);
10     LL a; cin >> a;
11     LL b = 1, c = 10, n;
12     if(a > 1 && a & 1) n = ( a - 1 ) / 2, b = 2 * n * n + 2 * n, c = b + 1;
13     else if(a > 4 && !(a & 1)) n = a / 2, b = n * n - 1, c = b + 2;
14     else if(a == 4) b = 3, c = 5;//特判
15     if(a != 1 && a != 2) cout << b << " " << c << endl;
16     else cout << "-1" << endl;
17     return 0;
18 }