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.
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.
Print the maximum possible number of blocks in a valid partitioning.
3 1 2 3
3
4 2 1 3 2
2
In the first sample the partitioning looks like that: [1][2][3].
In the second sample the partitioning is: [2, 1][3, 2]
给你 n 个数 arr[i],让你将这 n 个数分成 从小到大的顺序，看可以分成几块，当然，每个块中的可以排序。。。
样例解释：
3
1 2 3
可以分成 1 2 3三块，拍完许之后是从小到大的。。
解体思路：
将从 0 到 n-1 的每个dp值算出来，咋算呢，dp值就是从 i 到 n-1的最小值。。。
但是不要采用 两个 for 的算法容易超时，所以直接采用从后往前的算法，前边的每一个dp值分别跟 自己比，取最小的那一个。。
上代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e5+5;
const int mod = 1e9+7;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
LL gcd(LL a, LL b)
{
if(b == 0)
return a;
return gcd(b, a%b);
}
int arr[maxn];
int dp[maxn];
int main()
{
int n;
while(~scanf("%d",&n))
{
MM(dp);
MM(arr);
for(int i=0; i<n; i++)
scanf("%d",&arr[i]);
dp[n-1] = arr[n-1];
for(int i=n-2; i>=0; i--)
{
dp[i] = arr[i];
dp[i] = min(dp[i], dp[i+1]);
}
int ans = 1, Max = -INF;
for(int i=0; i<n; i++)
{
Max = max(Max, arr[i]);
if(dp[i+1] >= Max)
ans++;
}
cout<<ans<<endl;
}
return 0;
}
