Codeforces 599C Day at the Beach 【思维】

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.

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].

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

题意：给定n个元素的序列a[]，其升序序列为b[]。现在要求将序列a[]分成最多个连续的块，要求对每个块排序后的序列c[]与b[]相同。

思路：考虑最后结果得到的所有块，我们取出每个块的末尾元素构成一个升序序列d[top]。

注意到 (0 <= i < top) d[i] = b[i]。

我们求出a1 ... a[i]的最大值Max，ai ... a[n]的最小值Min。最后比较有多少个Min[i+1] >= Max[i]就可以了。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-4
#define MAXN (100000+10)
#define MAXM (1000000+100)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 100000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int Max[MAXN], a[MAXN];
int main()
{
	int n; Ri(n);
	Max[0] = 0;
	for(int i = 0; i < n; i++)
	{
		Ri(a[i]);
		Max[i+1] = max(Max[i], a[i]);
	}
	int r = INF, ans = 0;
	for(int i = n-1; i >= 0; i--)
	{
		r = min(r, a[i]);
		if(r >= Max[i])
            ans++;
	}
	Pi(ans);
	return 0;
}