Codeforces Round #262 (Div. 2) 460C. Present（二分）

题目链接：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

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.

题意：

给出N朵花的初始的高度。从左到右排列，最多浇水m天，每天仅仅能浇一次。每次能够使连续的 w 朵花的高度添加单位长度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：

从最低和最高（记得+m）的高度之间二分枚举高度，找出最大能适合的！见代码……

代码例如以下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL __int64
const int MAXN = 200017;
LL a[MAXN], b[MAXN], v[MAXN];
int main()
{
    LL n, m, w;
    while(~scanf("%I64d %I64d %I64d",&n,&m,&w))
    {
        LL low = 1e9, top = -1;
        for(int i = 1 ; i <= n ; i++)
        {
            scanf("%I64d", &a[i]);
            if(a[i] < low)
                low = a[i];
            if(a[i] > top)
                top = a[i];
        }
        top += m;//最大的高度
        LL mid, ans = -1 ;
        while(low <= top)
        {
            mid = (low + top)>>1 ;
            for(int i = 1 ; i <= n ; i++)
                b[i] = max(mid - a[i],(LL)0);//每朵花须要浇水的天数
            memset(v,0,sizeof(v));
            LL day = m;//天数
            LL c = 0;//已经浇了的天数
            for(int i = 1; i <= n; i++)
            {
                c += v[i];
                b[i] -= c;//已浇c天
                if(b[i] > 0)
                {
                    day -= b[i];
                    if(day < 0)//天数不够
                        break;
                    c += b[i];//已浇b[i]天
                    v[i+w] -= b[i];//浇水到这里
                    b[i] = 0;
                }
            }
            if(day < 0)//不符合，向更小的值二分寻找
                top = mid - 1;
            else//继续向更大的值二分寻找
            {
                ans = mid;
                low = mid + 1;
            }
        }
        printf("%I64d
", ans);
    }
    return 0;
}