Codeforces Round #262 (Div. 2) 460C. Present

题目连接：http://codeforces.com/problemset/problem/460/C

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

input

2 5 1
5 8

output

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.

题意：给出n朵花的初始的高度（a1, a2, ..., an），从左到右排列，最多浇水m天，每天只能浇一次，每次可以使连续的 w 朵花的高度增加单位长度1，问最后m天浇完水后最矮的花的高度最高是达到多少。

思路：从n朵花中最矮min{ai}和最高max{ai}+m之间二分枚举高度。

代码：

#include <iostream>
#include <cstdio>
#include <string.h>
#define MAX 100010
using namespace std;int n, m, w;  // n flowers, m days, w-width
int a[MAX];bool check(int mid)
{  // O(n)的时间处理 n-w+1 个宽为 w 的连续区间int grow[MAX];  // 每盆花的高度值int st[MAX];memset(grow,0,sizeof(grow));memset(st,0,sizeof(st));for (int i = 0; i < n; i++)grow[i] = a[i];int delta = 0;  int moves = m;for (int i = 0; i < n; i++){delta -= st[i];    // 除去对区间外浇花的影响grow[i] += delta;  // w-width区间内的花全部生长if (grow[i] < mid){int h = mid-grow[i];grow[i] = mid;delta += h;st[i+w<n?i+w:n] += h;  // O(1)时间moves -= h;if (moves < 0)return false;}}return true;
}int main()
{cin >> n >> m >> w;int left = 0x7fffffff, right = 0;memset(a,0,sizeof(a));for (int i = 0; i < n; i++){cin >> a[i];if (a[i] < left) left = a[i];if (a[i] > right) right = a[i];}right += m;int ans;while (left <= right){int mid = (left+right)>>1;if (check(mid)) {left = mid+1, ans = mid;} else right = mid-1;}cout << ans << endl;return 0;
}

最后想说的， check() 函数真的写得很巧妙，现在理解得也不是太透彻，是特别的数据结构吗？

每天一点点积累吧！

参考：

1. CodeforcesRound #262 (Div. 2) Editorial http://codeforces.ru/blog/entry/13465?locale=en

2. http://blog.csdn.net/accepthjp/article/details/38737429

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