Pythagorean Triples CodeForces - 707C

原题链接
考察:思维
思路:
  

\[a^2+b^2=c^2 \]

\[a^2 = c^2-b^2 = (c+b)(c-b) \]

\[令c+b = a^2 ,c-b = 1则 \]

\[c = \frac{a^2+1}{2} , b = \frac{a^2-1}{2} \]

  此时分5种情况:
(1) n = 1,无解
(2) n = 2,无解
(3) n为奇数,此时计算输出
(4) n为偶数,不断/2直到获得奇数.
(5) n只有2的因子,对应 3 4 5的情况.

Code

#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
int n;
void solve()
{
	if(n==1||n==2) {puts("-1");return;} 
	if(n&1)
	{
		LL a = (LL)n*n+1>>1;
		LL b = (LL)n*n-1>>1;
		printf("%lld %lld\n",b,a);
		return;
	}
	int cnt = 0;
	while(n%2==0)
	{
		n/=2;
		cnt++;
	}
	if(n==1)
	{
		cnt-=2;
		printf("%lld %lld\n",3*(1ll<<cnt),5*(1ll<<cnt));
		return;
	}
	LL a = ((LL)n*n+1>>1)*(1ll<<cnt);
	LL b = ((LL)n*n-1>>1)*(1ll<<cnt);
	printf("%lld %lld\n",b,a);
}
int main()
{
	scanf("%d",&n);
	solve();
	return 0;
}