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 plantednflowers 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 aremdays left to the birthday. The height of thei-th flower (assume that the flowers in the row are numbered from1tonfrom left to right) is equal toa_iat the moment. At each of the remainingmdays the beaver can take a special watering and waterwcontiguous 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 integersn,mandw(1 ≤ w ≤ n ≤ 10⁵; 1 ≤ m ≤ 10⁵). The second line contains space-separated integersa₁, a₂, ..., a_n(1 ≤ a_i ≤ 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.

二分+貪心，智商問題啊，當時就是沒想過二分，等比完，一聽說二分才想起來

二分可能的最大的可能高度，看能不能讓所有的高高度都超過mid，使用一個數組維護每一段的加減

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL __int64
LL a[210000] , b[210000] , c[210000] ;
int main()
{
    LL n , m , w , d , x , i;

    scanf("%I64d %I64d %I64d", &n, &m, &w);
    for(i = 1 ; i <= n ; i++)
        scanf("%I64d", &a[i]);
    LL low = 0 , mid , top = 1e10 , ans = -1 ;
    while(low <= top)
    {
        mid = (low + top)/2 ;
        for(i = 1 ; i <= n ; i++)
            b[i] = max(mid - a[i],(LL)0);
        memset(c,0,sizeof(c));
        d = m ; x = 0 ;
        for(i = 1 ; i <= n ; i++)
        {
            x += c[i] ;
            b[i] -= x ;
            if(b[i] > 0)
            {
                LL k ;
                d -= b[i] ;
                if(d < 0)
                    break;

                x += b[i] ;
                c[i+w] -= b[i] ;
                b[i] = 0 ;
            }
        }
        if(d < 0)
            top = mid - 1 ;
        else
        {
            ans = mid ;
            low = mid + 1 ;
        }
    }
    printf("%I64d\n", ans);
}