CodeForces 785C Anton and Fairy Tale

二分。

如果$n≤m$，显然只能$n$天。

如果$n>m$，至少可以$m$天，剩余还可以支撑多少天，可以二分计算得到，也可以推公式。二分计算的话可能爆$long$ $long$，上了个$Java$。

import java.math.BigInteger;
import java.util.Scanner;
 
public class Main
{
	static Scanner cin = new Scanner(System.in);
 
	static BigInteger sum(BigInteger L,BigInteger R)
	{
		if(L.compareTo(R)>0) return BigInteger.ZERO;
		BigInteger A = L.add(R);
		BigInteger B = R.subtract(L); B = B.add(BigInteger.ONE);
		BigInteger c = A.multiply(B);
		return c.divide(BigInteger.valueOf(2));
	}
 
	public static void main(String []args)
	{
		BigInteger m,n;
 
		n = cin.nextBigInteger();
		m = cin.nextBigInteger();
 
		if(n.compareTo(m)<=0)
		{
			System.out.println(n);
		}
 
		else
		{
			BigInteger ans = m;
			BigInteger sy = n.subtract(m);
 
			BigInteger hai = BigInteger.ZERO;
 
			BigInteger L = BigInteger.ZERO, R = sy;
 
			boolean pp=false;
			while(L.compareTo(R)<=0)
			{
				BigInteger mid = (L.add(R).divide(BigInteger.valueOf(2)));
				if(sum(ans.add(BigInteger.ONE),ans.add(mid)).compareTo(sy.add(m.multiply(mid)))<0)
				{
					pp=true;
					hai = mid;
					L = mid.add(BigInteger.ONE);
				}
 
				else if(sum(ans.add(BigInteger.ONE),ans.add(mid)).compareTo(sy.add(m.multiply(mid)))==0)
				{
					pp=false;
					hai = mid;
					L = mid.add(BigInteger.ONE);
				}
 
				else R = mid.subtract(BigInteger.ONE);
			}
 
			ans = ans.add(hai);
 
			if(pp == true) ans = ans.add(BigInteger.ONE);
			System.out.println(ans);
 
		}
 
	}
}