题目大意：给你一行数字，每个数字代表这个位置的花初始高度，总共有m天，每天可以给连续的w朵花浇水一次，之后，浇过水的花就会长高单位1，问，最后，最矮的花最高有多高。

思路：用二分枚举所有可能的高度，如果达到这个高度需要的天数小于m就增加下限，否则缩小上限。

原题：

C. Present

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his informatics teacher is going to have a birthday and the beaver has decided to prepare a present for her. He planted n flowers in a row on his windowsill and started waiting for them to grow. However, after some time the beaver noticed that the flowers stopped growing. The beaver thinks it is bad manners to present little flowers. So he decided to come up with some solutions.

There are m days left to the birthday. The height of the i-th flower (assume that the flowers in the row are numbered from 1 to n from left to right) is equal to ai at the moment. At each of the remaining m days the beaver can take a special watering and water w contiguous flowers (he can do that only once at a day). At that each watered flower grows by one height unit on that day. The beaver wants the height of the smallest flower be as large as possible in the end. What maximum height of the smallest flower can he get?

Input

The first line contains space-separated integers n, m and w (1?≤?w?≤?n?≤?105; 1?≤?m?≤?105). The second line contains space-separated integers a1,?a2,?...,?an (1?≤?ai?≤?109).

Output

Print a single integer — the maximum final height of the smallest flower.

Sample test(s)

Input
6 2 3
2 2 2 2 1 1

Output
2

Input
2 5 1
5 8

Output
9

Note

In the first sample beaver can water the last 3 flowers at the first day. On the next day he may not to water flowers at all. In the end he will get the following heights: [2, 2, 2, 3, 2, 2]. The smallest flower has height equal to 2. It‘s impossible to get height 3 in this test.

我的代码如下：

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
long long n,m,w,aa[100010];
void in()
{
	long long i;
	cin>>n>>m>>w;
	for(i=0;i<n;i++)
		cin>>aa[i];
}
void work()
{
	long long tt,tmp[200010],cnt,t,i,l=1,r=1000100000,x;
	while(l<r)
	{
		tt=cnt=0;
		memset(tmp,0,sizeof(tmp));
		x=(l+r+1)>>1;
		for(i=0;i<n;i++)
		{
			cnt-=tmp[i];
			t=x-aa[i]-cnt;
			if(t>0)
			{
				tt+=t;
				cnt+=t;
				tmp[i+w]=t;
			}
		}
		if(tt<=m)
			l=x;
		else
			r=x-1;
	}
	cout<<l;
}
int main()
{
	in();
	work();
}

二分解Codeforces Round #262 (Div. 2)C. Present