Codeforces Round #332 (Div. 2) C. Day at the Beach-优快云博客

一场海滩上的建造比赛激发了Squidward、Spongebob和Patrick的创造力，他们建造了多个沙堡。面对混乱的沙堡排列，他们决定通过将沙堡分成若干组并独立排序每组来解决这个问题。本文介绍了一种有效的算法来确定最大数量的组数，同时确保整体序列有序。

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One day Squidward, Spongebob and Patrick decided to go to the beach. Unfortunately, the weather was bad, so the friends were unable to ride waves. However, they decided to spent their time building sand castles.

At the end of the day there were n castles built by friends. Castles are numbered from 1 to n, and the height of the i-th castle is equal to hi. When friends were about to leave, Squidward noticed, that castles are not ordered by their height, and this looks ugly. Now friends are going to reorder the castles in a way to obtain that condition hi ≤ hi + 1 holds for all i from 1 to n - 1.

Squidward suggested the following process of sorting castles:

Castles are split into blocks — groups of consecutive castles. Therefore the block from i to jwill include castles i, i + 1, ..., j. A block may consist of a single castle.
The partitioning is chosen in such a way that every castle is a part of exactly one block.
Each block is sorted independently from other blocks, that is the sequence hi, hi + 1, ..., hjbecomes sorted.
The partitioning should satisfy the condition that after each block is sorted, the sequence hibecomes sorted too. This may always be achieved by saying that the whole sequence is a single block.

Even Patrick understands that increasing the number of blocks in partitioning will ease the sorting process. Now friends ask you to count the maximum possible number of blocks in a partitioning that satisfies all the above requirements.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of castles Spongebob, Patrick and Squidward made from sand during the day.

The next line contains n integers hi (1 ≤ hi ≤ 109). The i-th of these integers corresponds to the height of the i-th castle.

Output

Print the maximum possible number of blocks in a valid partitioning.

   Sample test(s)
 
     input
   
3
1 2 3

     output
   
3

     input
   
4
2 1 3 2

     output
   
2

Note

In the first sample the partitioning looks like that: [1][2][3].

$\text{[math]}$

In the second sample the partitioning is: [2, 1][3, 2]

$\text{[math]}$

解题报告：Editorial Codeforces Round #332 (Div. 2)

我的代码

#include
   
    
#include
    
     
#include
     
      
#include
      
       
using namespace std;
const int N=100005;
int h[N],a[N];
int kk;
int erfen(int num)    ///二分
{
    int l=0,r=kk-1,mid=(0+kk-1)/2;
    if(a[l]>num) return 0;
    if(a[r]<=num) return r+1;
    while(l
       
        num) r=mid;
        mid=(l+r)/2;
        if(a[mid]<=num&&a[mid+1]>num) return mid+1;
    }
}
int main()
{
    int i,n,maxn;
    scanf("%d",&n);
    for(i=0; i
        
         h[i]?maxn:h[i]; if(a[kk-1]<=h[i]) { a[kk]=h[i]; ///a[kk]存各分块的最大值 } else ///小于 二分查找下一分块的位置 { kk=erfen(h[i]); a[kk]=maxn; ///a[kk]存各分块的最大值 } kk++; } printf("%d\n",kk); return 0; }

大神的

#include 
  
   
using namespace std;
const int N=100002;
int n,h[N],i,c,s[N],m;
int main()
{
    scanf("%d",&n);
    s[n-1]=1e9+1;
    for(i=0; i
   
    =0; i--)
    {
        s[i]=min(s[i+1],h[i+1]);
    }
    for(i=0; i
    
     =m)c++;
    }
    printf("%d\n",c);
    return 0;
}