D. Pythagorean Triples (math、暴力)

题目

题解:
D. Pythagorean Triples (math、暴力)_第1张图片因此c=(a*a-1)/2+1
我们暴力枚举a,从a=1到(a * a-1)/2+1<=n (C需要<=n)
如果满足(a * a-1)%2==0,即b为整数,且c=b+1,C也为整数,符合条件,贡献+1

Code:

#include
#include
#define pii pair
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;

int main()
{
     
	FAST;
	int t;cin >> t;
	while (t--)
	{
     
		ll n;cin >> n;
		int sum = 0;
		for (ll i = 1;((i * i) - 1) / 2 + 1 <= n;i++)
		{
     
			ll b = i * i - 1;
			if (b % 2 == 0 && b >= i) sum++;
		}
		cout << sum << endl;
	}
}

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