Low Elements--AT

题目描述

Given is a permutation P1,…,PN of 1,…,N. Find the number of integers i (1≤i≤N) that satisfy the following condition:
·For any integer j (1≤j≤i), Pi≤Pj.
Constraints
·1≤N≤2×105
·P1,…,PN is a permutation of 1,…,N.
·All values in input are integers.

输入

Input is given from Standard Input in the following format:
N
P1 … PN

输出

Print the number of integers i that satisfy the condition.

样例输入 Copy

【样例1】
5
4 2 5 1 3
【样例2】
4
4 3 2 1
【样例3】
6
1 2 3 4 5 6
【样例4】
8
5 7 4 2 6 8 1 3
【样例5】
1
1

样例输出 Copy

【样例1】
3
【样例2】
4
【样例3】
1
【样例4】
4
【样例5】
1

提示

样例1解释
i=1, 2, and 4 satisfy the condition, but i=3 does not - for example, Pi>Pj holds for j=1.
Similarly, i=5 does not satisfy the condition, either. Thus, there are three integers that satisfy the condition.
样例2解释
All integers i (1≤i≤N) satisfy the condition.
样例3解释
Only i=1 satisfies the condition.

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
int cnt;
start{
    int n=read;
    int minn=mod;
    for(int i=1;i<=n;i++){
        int num=read;
        minn=min(num,minn);
        if(minn==num)
            cnt++;
    }
    cout<<cnt<<endl;
    end;
}
 
/**************************************************************
    Language: C++
    Result: 正确
    Time:24 ms
    Memory:2024 kb
****************************************************************/