Codeforces 985 D. Sand Fortress (二分)

Description

You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.

Obviously, there is not enough sand on the beach, so you brought $n$ packs of sand with you. Let height $h_i$ of the sand pillar on some spot $i$ be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with $H$ sand packs to the left of the first spot and you should prevent sand from going over it.

Finally you ended up with the following conditions to building the castle:

  • $h_1 ≤ H$: no sand from the leftmost spot should go over the fence;
  • For any $i\in[1,∞)]$ $|h_i - h_{i + 1}| ≤ 1$: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
  • $\sum_{i=1}^{∞}h_i=n$: you want to spend all the sand you brought with you.

As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.

Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.

 

Input

The only line contains two integer numbers $n$ and $H~(1 ≤ n, H ≤ 10^{18})$ — the number of sand packs you have and the height of the fence, respectively.

 

Output

Print the minimum number of spots you can occupy so the all the castle building conditions hold.

 

Examples input

5 2

 

Examples output

3

 

题意

有 $n$ 块沙包,左侧挡板的高度为 $H$ ,若某一格的沙包个数与相邻格的沙包个数之差大于 1,则该格内的沙包可能会发生滑落,因此这样的情况是不被允许的,现要你求出最少需要多少格才能完全放置所有的沙包。

 

思路

显然最后放置好沙包以后的图形从左到右大体趋势要么一直递减,要么先上升后递减,于是我们二分求临界点即可。

 

AC 代码

def judge(mid, n, H):
    if mid <= H:  # 一直递减的情形
        return mid * (mid + 1) // 2

    # 左右两侧都是等差数列
    right = mid * (mid + 1) // 2
    left = (mid - H + 1) * H + (mid - H + 1) * ((mid - H + 1) - 1) // 2
    return left + right - mid


def solve(n, H):
    low = 1
    high = 1000000000000000000
    while True:
        mid = (low + high) >> 1
        now = judge(mid, n, H)
        if low == mid:
            if now >= n:
                return low + max(0, low - H)
            elif now + low >= n:  # 能否再加一列使其满足条件
                return low + max(0, low - H) + 1
            return high + max(0, high - H)
        if now == n:
            return mid + max(0, mid - H)
        elif now < n:
            low = mid
        elif now > n:
            high = mid
    pass


try:
    while True:
        n, H = input().split()
        n = int(n)
        H = int(H)
        print(solve(n, H))
except EOFError:
    pass