Educational Codeforces Round 52 Editorial B

简单的模拟题
就是一开始没看懂标程

标程：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n;
    long long m;
    cin >> n >> m;
    long long cur = 1;
    long long rem = m;
    
    //表示的是大于m的最小符合n*（n-1）/2的n
    while(rem > 0){
        long long d = min(cur, rem);
        rem -= d;
        ++cur;
    }
    
    assert(rem == 0);
    
    long long res = n;
    if(cur > 1) res = n - cur;
    
    cout << max(0ll, n - m * 2) << ' ' << res << endl;
    return 0;
}

法二：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll n, m;

int main()
{
	cin >> n >> m;
	cout << max(n-2*m,0LL) << " ";
	if(m == 0)
    {
        cout << n << endl;
        return 0;
    }
    for(ll i = 1; i <= n; i++)
    {
        if(m <= i*(i-1)/2)
        {
            cout << n-i << endl;
            return 0;
        }
    }
    return 0;
}