CF#262(div2) C——Present（二分）

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我是靠谱客的博主默默机器猫，最近开发中收集的这篇文章主要介绍CF#262(div2) C——Present（二分），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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.

唉，比赛的时候居然没有做出来！！！

二分可以想到，关键是没有找到好的方法验证可行性。

其实应该用扫描线的思想，O(n)即可。

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <vector>
#define INF 0x7fffffff
using namespace std;
typedef long long LL;
int a[100005];
int f[100005];
int n,m,w;
bool check(int x)
{
memset(f,0,sizeof(f));
int k=m;
for(int i=0;i<n;i++)
{
if(i!=0) f[i]+=f[i-1];
int d=max(0,x-a[i]-f[i]);
f[i]+=d;
f[min(n,i+w)]-=d;
k-=d;
if(k<0)
return false;
}
return true;
}
int main()
{
scanf("%d%d%d",&n,&m,&w);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
long long l=1,r=INF,mid;
while(l<r)
{
mid=(l+r+1)/2;
if(check(mid))
l=mid;
else
r=mid-1;
}
printf("%I64dn",l);
return 0;
}