D. Pythagorean Triples (math、暴力)

题目

题解：
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<iostream>
#include<cmath>
#define pii pair<int,int>
#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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