Codeforces Round #404 (Div. 2) C 二分查找

Posted 2020-09-07 yoyo_sincerely

Codeforces Round #404 (Div. 2)

题意：对于 n and m (1?≤?n,?m?≤?10^18)  找到

　　1) [n<= m] cout<<n;

　　2) [n>m]最小的 k => (k -m) * (k-m+1) >= (n-m)*2 成立

思路：二分搜索

	#include <bits/stdc++.h>
	#include <map>
	using namespace std;
	
	#define LL long long
	const long long INF = 1e10;
	long long n, m, del, mix;
	bool cal(long long a){
		return a*(a+1) >= del; 
	}
	
	long long find(long long l, long long r){
		while(l + 1 < r){
			long long mid = (l +r) >> 1;
//			cout<<(l+r)/2<<" "<<l<<" "<<r<<endl;
			if(cal(mid)) r = mid;
			else l = mid;
		}
		return l;
	}
	
	int main(){
		ios::sync_with_stdio(false); cin.tie(0);
//		freopen("input2.txt", "r", stdin);
		cin>>n>>m;
		
		if(n <= m) cout<<n<<endl;
		if(n > m){
			del = 2 * (n-m);
			mix = find(0, 1e10);
			while(!cal(mix)) mix++;
			cout<<mix+m<<endl;
		}
		return 0;
	}