codeforces 551 C GukiZ hates Boxes

……睡太晚了。

。。脑子就傻了……

这个题想的时候并没有想到该这样……

题意大概是有n堆箱子从左往右依次排列，每堆ai个箱子，有m个人，最開始都站在第一个箱子的左边，

每个人在每一秒钟都必须做出两种选择中的一种：1若他的位置有箱子则搬走一个箱子，2往右走一步。

问把全部箱子都搞掉的最少时间……

非常显然二分一下答案，若为x秒，则每一个人都有x秒。一个一个排出去搬。看是否可以搬完……

我居然没想到……

#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.