let mpat 沒有想到 做出 word sid other ide arr

……睡太晚了。

。。腦子就傻了……

這個題想的時候並沒有想到該這樣……

題意大概是有n堆箱子從左往右依次排列，每堆ai個箱子，有m個人，最開始都站在第一個箱子的左邊，

每個人在每一秒鐘都必須做出兩種選擇中的一種：1若他的位置有箱子則搬走一個箱子，2往右走一步。

問把全部箱子都搞掉的最少時間……

非常顯然二分一下答案，若為x秒，則每一個人都有x秒。一個一個排出去搬。看是否可以搬完……

我居然沒想到……

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
int a[100010];
bool isok(int n,int m,ll x)
{
	ll t=x;
	for(int i=0;i<n;i++)
	{
		int re=a[i];
		ll t1=t-i-1;
		if(t1<0)
		{
			if(m==0)
				return 0;
			t1=x-i-1;
			t=x;
			m--;
		}
		if(re>0)
		{
			if(t1>=re)
				t-=re;
			else
			{
				re-=t1;
				t1=x-i-1;
				t=x;
				int t2=re/t1;
				re-=t2*t1;
				if(re>0)
				{
					t2++;
					t-=re;
				}
				else
					t=0;
				m-=t2;
				if(m<0)
					return 0;
			}
		}
	}
	return 1;
}
int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=0;i<n;i++)
		cin>>a[i];
	for(int i=n-1;i>-1;i--)
		if(a[i]!=0)
		{
			n=i+1;
			break;
		}
	ll l=n+1,r=ll(1e15);
	while(l<=r)
	{
		ll md=l+r>>1;
		if(isok(n,m-1,md))
			r=md-1;
		else
			l=md+1;
	}
	cout<<l;
}

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.

In total there are n

piles of boxes, arranged in a line, from left to right, i-th pile (1?≤?i?≤?n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:

1. If i?≠?n, move from pile i to pile i?+?1;
2. If pile located at the position of student is not empty, remove one box from it.

GukiZ‘s students aren‘t smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn‘t want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ‘s way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.

Input

The first line contains two integers n and m (1?≤?n,?m?≤?105), the number of piles of boxes and the number of GukiZ‘s students.

The second line contains n integers a1,?a2,?... an (0?≤?ai?≤?109) where ai represents the number of boxes on i-th pile. It‘s guaranteed that at least one pile of is non-empty.

Output

In a single line, print one number, minimum time needed to remove all the boxes in seconds.

Sample test(s) input

2 1
1 1

output

4

input

3 2
1 0 2

output

5

input

4 100
3 4 5 4

output

5

Note

First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).

Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.

Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.

codeforces 551 C GukiZ hates Boxes