cf div2(262)c题

最新推荐文章于 2024-09-03 14:51:25 发布

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通过二分查找的方法，在限定的操作次数内，找到能使所有花达到的最大最小高度。

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.

题目大意：给你n多花，进行m次操作，每次操作使连续的w多花高度加一，求m次操作后使得最矮花高度最大，输出他的高度

解题思路：首先我们二分可能到达的高度，让后check是否所有花能到大吃高度，如果能，我们继续找更大的，如果不能，就找更小的。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define inf  1000100000
#define N 100005
#define LL long long

int n,m,w;
int a[N],f[N];
bool check(int x)
{
    int k=m,i,d;
    memset(f,0,sizeof(f));
    for(i=1;i<=n;i++)
    {
        f[i]+=f[i-1];
        d=max(0,x-a[i]-f[i]);
        k-=d;
        f[i]+=d;
        f[min(i+w,n+1)]-=d;
        if(k<0)
            return false;
    }
    return true;
}
int main()
{
    int i;
    while(scanf("%d%d%d",&n,&m,&w)!=EOF)
    {
        LL mi=inf;
        LL ma=-inf;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            mi=min(mi,(LL)a[i]);
            ma=max(ma,(LL)a[i]);
        }
        LL l=mi,r=ma+m;
        LL mid,ans;
        while(l<=r)
        {
            mid=(l+r)/2;
            if(check(mid))
            {
                l=mid+1;
                ans=mid;
            }
            else
                r=mid-1;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}