二分。

如果$n≤m$，顯然只能$n$天。

如果$n>m$，至少可以$m$天，剩余還可以支撐多少天，可以二分計算得到，也可以推公式。二分計算的話可能爆$long$ $long$，上了個$Java$。

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);
			
		}
		
	}
}

CodeForces 785C Anton and Fairy Tale