Problem:
Score : 300 points

Problem Statement

Given is a permutation P_1, \ldots, P_N of 1, \ldots, N.
Find the number of integers i (1 \leq i \leq N) that satisfy the following condition:

For any integer j (1 \leq j \leq i), P_i \leq P_j.

Constraints

1 \leq N \leq 2 \times 10^5

P_1, \ldots, P_N is a permutation of 1, \ldots, N.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1 ... P_N

Output

Print the number of integers i that satisfy the condition.

Sample Input 1

5
4 2 5 1 3

Sample Output 1

3

i=1, 2, and 4 satisfy the condition, but i=3 does not - for example, P_i > P_j holds for j = 1.

Similarly, i=5 does not satisfy the condition, either. Thus, there are three integers that satisfy the condition.

Sample Input 2

4
4 3 2 1

Sample Output 2

4

All integers i (1 \leq i \leq N) satisfy the condition.

Sample Input 3

6
1 2 3 4 5 6

Sample Output 3

1

Only i=1 satisfies the condition.

Sample Input 4

8
5 7 4 2 6 8 1 3

Sample Output 4

4

Sample Input 5

1
1

Sample Output 5

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
#include <string>
#include <queue>
#include <cmath>
#include <iomanip>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <numeric>
#define ll long long
#define fr(i,k,N) for(decltype(N+k) i=k; i<N;i++)
#define rep(i,N) for(decltype(N) i=0; i<N;i++)
#define ALL(v) v.begin(),v.end()
using namespace std;
template <class X> void pr(X test){cout<<test<<endl;}
template <class X> void prr(X test){for(auto it:test){cout<<it<< " ";}cout<<endl;}
template <class X> void prrr(X test){int f=0;for(auto it:test){cout<<(f++!=0?" ":"")<<it;}cout<<endl;}
template<typename X>
istream&operator>>(istream&i,vector<X>&v){for(X&x:v)i>>x;return i;}

class UF { // union find from https://github.com/kartikkukreja/blog-codes (MIT)
    // [usage]
    // auto uf = new UF(n);
    // uf->merge(l, r); // 0 <= l, r <= n-1
    // ...
public:
    int *id; // inner id
    int cnt; // number of all vartices
    int *sz; // sz[i] means size of i's set
    UF(int N)   {
        cnt = N;
        id = new int[N];
        sz = new int[N];
        for(int i=0; i<N; i++) {
            id[i] = i;
            sz[i] = 1;
        }
    }
    ~UF() {
        delete [] id;
        delete [] sz;
    }
    int find(int p) {
        int root = p;
        while (root != id[root]) root = id[root];
        while (p != root) {
            int newp = id[p];
            id[p] = root;
            p = newp;
        }
        return root;
    }
    void merge(int x, int y) {
        int i = find(x);
        int j = find(y);
        if (i == j) return;
        
        if (sz[i] < sz[j]) {
            id[i] = j;
            sz[j] += sz[i];
        } else {
            id[j] = i;
            sz[i] += sz[j];
        }
        cnt--;
    }
    bool connected(int x, int y) {
        return find(x) == find(y);
    }
    int count() {
        return cnt;
    }
};

// return 約数unordered_set 1 <= x <= n
auto divv = [](auto n) {
  unordered_set<decltype(n)> v;
  for(decltype(n) i = 1; i*i <= n; i++) {
    if(n%i == 0) {
      v.insert(i);
      if(i*i != n) v.insert(n/i);
    }
  }
  return v;
};

// check prime (and n>=2) or not
auto isPrime = [](auto n){
  if(n <= 1) return false;
  for(decltype(n) i = 2; i <= sqrt(n); ++i) {
      if(n % i == 0) {
          return false;
      }
  }
  return true;
};

// 最小公倍数
auto lcm = [](auto a, auto b){
    if(a<b) swap(a,b);
    auto r = a % b;
    auto x = a*b;
    while(r!=0){
        a = b;
        b = r;
        r = a % b;
    }
    return x/b;
};

// 最大公約数
auto gcd = [](auto a, auto b){
    return (a * b)/lcm(a,b);
};

// factorial n mod prm
ll factmod(ll a, ll prm){
    return a == 0 ? 1 : (a * factmod(a-1, prm)) % prm;
};

// 1/x mod prm
auto invmod = [](auto x, auto prm){
    decltype(x) r = 1;
    if (x==1) return r;
    for (auto ex = prm - 2; ex; ex >>= 1) {
        if (ex&1) r = (r * x) % prm;
        x = (x * x) % prm;
    }
    return r;
};

int main() {
  ll n;
  cin>>n;
  vector<ll> p(n);
  cin>>p;
  ll mm = p[0];ll sum = 0;
  rep(i, n){
      mm = min(mm, p[i]);
      if(p[i] == mm){
          sum++;
      }
  }
  pr(sum);
  
}