Codeforces Round #332 (Div. 2)-C. Day at the Beach

原题链接

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 tohi.
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 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 hi, hi + 1, ..., hj becomes
sorted.

The partitioning should satisfy the condition that after each block is sorted, the sequence hi 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 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.

Examples

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].

定义一个结构体Node,其中有两边变量v,i,   v表示值,i表示v在原来数列中处在第几位。再对node数组按照v从小到大排序，如果v相同则按照i从小到大排序。int l = 0, r = node[0].r;遍历l, 到r中的每个node[i].i观察其是否在[l, r]中，若不是更新r，最终使得[l, r]中每个node[i].i在该区间内，则为一个valid
partitioning，ans++.以此类推.

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <ctime>
#define maxn 100005
#define INF 1e18
using namespace std;
typedef long long ll;

struct Node{
int v, i;
friend bool operator < (const Node &a, const Node &b){
return a.v < b.v || (a.v == b.v && a.i < b.i);
}
}node[maxn];
int main(){

//	freopen("in.txt", "r", stdin);
int n;

scanf("%d", &n);
for(int i = 0; i < n; i++){
scanf("%d", &node[i].v);
node[i].i = i;
}
sort(node, node+n);
int ans = 0;
int l = 0, r = node[0].i;
while(1){
ans++;
for(int i = l+1; i <= r; i++){
r = max(r, node[i].i);
}
if(r == n-1)
break;
l = r + 1;
r = node[l].i;
}
printf("%d\n", ans);
return 0;
}