C. Day at the Beach
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 h__i. 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 h__i ≤ h__i + 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 j will 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 h__i, h__i + 1, …, h__j becomes sorted.
- The partitioning should satisfy the condition that after each block is sorted, the sequence h__i becomes 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 h__i (1 ≤ h__i ≤ 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
代码实现
1/*************************************************************************
2> File Name: code/cf/#332/C.cpp
3> Author: 111qqz
4> Email: rkz2013@126.com
5> Created Time: 2015年11月21日 星期六 03时14分55秒
6************************************************************************/
7
8#include<iostream>
9#include<iomanip>
10#include<cstdio>
11#include<algorithm>
12#include<cmath>
13#include<cstring>
14#include<string>
15#include<map>
16#include<set>
17#include<queue>
18#include<vector>
19#include<stack>
20#include<cctype>
21#define fst first
22#define sec second
23#define lson l,m,rt<<1
24#define rson m+1,r,rt<<1|1
25#define ms(a,x) memset(a,x,sizeof(a))
26using namespace std;
27const double eps = 1E-8;
28const int dx4[4]={1,0,0,-1};
29const int dy4[4]={0,-1,1,0};
30typedef long long LL;
31const int inf = 0x3f3f3f3f;
32const int N=1E5+7;
33int n;
34int a[N];
35int l[N],r[N];
36int main()
37{
38#ifndef ONLINE_JUDGE
39freopen("in.txt","r",stdin);
40#endif
41cin>>n;
42for ( int i = 1 ; i <= n ; i++) scanf("%d",&a[i]);
43r[n+1] = inf;
44for ( int i = n ; i >= 0 ; i --) r[i] = min(r[i+1],a[i]);
45int mx = -1;
46int cnt = 0 ;
47for ( int i = 1 ; i <= n ; i++)
48{
49mx = max(mx,a[i]);
50if (mx<=r[i+1])
51cnt++,mx = -1;
52}
53printf("%dn",cnt);
54
55
56#ifndef ONLINE_JUDGE
57#endif
58fclose(stdin);
59return 0;
60}