Codeforces 707C Pythagorean Triples

链接

传送门

题意

给出\(n(1 \leq n \leq 10^9)\)，输出与\(n\)能够组成直角三角形的两条边的长度，无解输出－1。

思路

• \(n \leq 2\)时，无解。
• \(n \geq 3\)且为奇数时，满足\({\left( \frac{n^2 - 1}{2} \right)}^2 + n^2 = {\left( \frac{n^2 + 1}{2} \right)}^2\)
• \(n \geq 4\)且为偶数时，满足\({\left( \frac{n^2}{4} - 1 \right)}^2 + n^2 = {\left( \frac{n^2}{4} + 1 \right)}^2\)

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

int main() {
  int n;
  scanf("%d", &n);
  if (n <= 2) {
    puts("-1");
    return 0;
  }
  LL m = 0, k = 0;
  if (n & 1) {
    m = (LL(n) * n - 1) / 2;
    k = m + 1;
  } else {
    m = LL(n) * n / 4 - 1;
    k = m + 2;
  }
  printf("%I64d %I64d\n", m, k);
  return 0;
}