C. GukiZ hates Boxes
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/551/problem/C
Description
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 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=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** ll a[maxn],b[maxn];
int n,m;
bool check(ll x)
{
for(int i=;i<=n;i++)
b[i]=a[i];
ll tmp=;
int cur=n;
for(int i=;i<=m;i++)
{
while(cur>=&&b[cur]==)
cur--;
if(cur==)
return ;
tmp=cur;
if(tmp>x)
continue;
while(cur>=&&b[cur]+tmp<=x)
{
tmp+=b[cur];
b[cur]=;
cur--;
}
if(cur==)
return ;
b[cur]-=(x-tmp);
}
return ;
}
int main()
{
//test;
ll sum=;
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)
{
scanf("%lld",&a[i]);
sum+=a[i];
}
sum+=n;
ll L=,R=sum;
while(L<R)
{
ll M=(L+R)>>;
if(check(M))
R=M;
else
L=M+;
}
printf("%lld\n",R);
return ;
}