Codeforces Round #307 (Div. 2) C. GukiZ hates Boxes 贪心/二分

C. GukiZ hates Boxes

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/551/problem/C

Description

Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.

In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:

    If i ≠ n, move from pile i to pile i + 1;
    If pile located at the position of student is not empty, remove one box from it.

GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.

The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.

Output

In a single line, print one number, minimum time needed to remove all the boxes in seconds.

Sample Input

2 1
1 1

Sample Output

4

HINT

 

题意

有一群小学生帮着老师搬东西，一开始小学生都在0点，然后每次每个小学生有两种操作，1是移动一格，2是搬走一堆东西

然后问你花费最少的时间

题解：

想dp的都是傻逼（没错就是我！

直接二分答案之后，然后贪心就好了……

代码：

 

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 2000001
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************


ll a[maxn],b[maxn];
int n,m;
bool check(ll x)
{
    for(int i=1;i<=n;i++)
        b[i]=a[i];
    ll tmp=0;
    int cur=n;
    for(int i=1;i<=m;i++)
    {
        while(cur>=1&&b[cur]==0)
            cur--;
        if(cur==0)
            return 1;
        tmp=cur;
        if(tmp>x)
            continue;
        while(cur>=1&&b[cur]+tmp<=x)
        {
            tmp+=b[cur];
            b[cur]=0;
            cur--;
        }
        if(cur==0)
            return 1;
        b[cur]-=(x-tmp);
    }
    return 0;
}
int main()
{
    //test;
    ll sum=0;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        sum+=a[i];
    }
    sum+=n;
    ll L=0,R=sum;
    while(L<R)
    {
        ll M=(L+R)>>1;
        if(check(M))
            R=M;
        else
            L=M+1;
    }
    printf("%lld\n",R);
    return 0;
}