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 (Clang 3.8.0) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<queue>
#include<stack>
#include<bitset>
#include<functional>
#include<map>
#include<iomanip>
#include<limits>
#include<unordered_set> 
#include<cmath>
#include <numeric>
#include <array>
#include <complex>
using namespace std;
long long p = 998244353;
//long long p = 1000000007;
#define int long long
#define vel vector<long long>
#define vvel vector<vel>
#define rep(i,n) for(int i=0;i<n;i++)
#define sor(v) sort(v.begin(),v.end())
#define mmax(a,b) a=max(a,b)
#define mmin(a,b) a=min(a,b)
#define mkp make_pair
#define pin pair<int,int>
#define qin pair<pin,int>
#define V vector
#define Endl endl
#define veb vector<bool>
#define fcout cout << fixed << setprecision(15)
#define rev(s) reverse(s.begin(),s.end())
#define lower(h,val) lower_bound(h.begin(),h.end(),val)-h.begin()
#define upper(h,val) upper_bound(h.begin(),h.end(),val)-h.begin()
vel kai;
vel inv_kai;
int rui(int a, int n, int mod) {
    if (n == 0) { return 1 % mod; }
    int x = rui(a, n / 2, mod);
    x *= x; x %= mod;
    if (n % 2 == 1) { x *= a; x %= mod; }
    return x;
}
int root(int x, vel& pa) {
    if (pa[x] == -1) { return x; }
    int ans = root(pa[x], pa); pa[x] = ans;
    return ans;
}
bool mar(int x, int y, vel& pa) {
    x = root(x, pa);
    y = root(y, pa);
    if (x != y) { pa[x] = y; }
    return (x != y);
}
int gcd(int x, int y) {
    if (x < y) { return gcd(y, x); }
    if (y == 0) { return x; }
    return gcd(y, x % y);
}
int lcm(int x, int y) {
    x = abs(x); y = abs(y);
    return x * y / gcd(x, y);
}
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
void make_kai(int max_kai) {
    kai = vel(max_kai, 1);
    inv_kai = kai;
    rep(i, max_kai - 1) {
        kai[i + 1] = kai[i] * (i + 1); kai[i + 1] %= p;
        inv_kai[i + 1] = modinv(kai[i + 1], p);
    }
}
int com(int n, int r) {
    if ((n < 0) || (r < 0) || (r > n)) { return 0; }
    int ans = (kai[n] * inv_kai[r]) % p;
    return (ans * inv_kai[n - r]) % p;
}
vel uni(vel x) {
    if (x.size() == 0) { return x; }
    sor(x);
    int n = x.size();
    vel ans(1, x[0]);
    for (int j = 1; j < n; j++) {
        if (x[j - 1] != x[j]) { ans.push_back(x[j]); }
    }
    x = ans;
    return x;
}
void pr(vel& v) {
    int n = v.size();
    if (n != 0) {
        cout << v[0];
        rep(i, n - 1) {
            cout << " " << v[i + 1];
        }
        cout << endl;
    }
}
vel dijk(V<V<pin>>& way, int st, int inf) {
    int n = way.size();
    vel dist(n, inf); dist[st] = 0;
    priority_queue<pin, vector<pin>, greater<pin>> pq;
    pq.push(mkp(0, st));
    veb is_checked(n, false);
    while (!pq.empty()) {
        pin x = pq.top(); pq.pop();
        int pot = x.second;
        if (!is_checked[pot]) {
            is_checked[pot] = true;
            for (auto y : way[pot]) {
                int nex_dist = x.first + y.second;
                int nex_pot = y.first;
                if (dist[nex_pot] > nex_dist) {
                    dist[nex_pot] = nex_dist;
                    pq.push(mkp(nex_dist, y.first));
                }
            }
        }
    }
    return dist;
}
template< typename Monoid >
struct SegmentTree {
    using F = function< Monoid(Monoid, Monoid) >;

    int sz;
    vector< Monoid > seg;

    const F f;
    const Monoid M1;

    SegmentTree(int n, const F f, const Monoid& M1) : f(f), M1(M1) {
        sz = 1;
        while (sz < n) sz <<= 1;
        seg.assign(2 * sz, M1);
    }

    void set(int k, const Monoid& x) {
        seg[k + sz] = x;
    }

    void build() {
        for (int k = sz - 1; k > 0; k--) {
            seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
        }
    }

    void update(int k, const Monoid& x) {
        k += sz;
        seg[k] = x;
        while (k >>= 1) {
            seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
        }
    }

    Monoid query(int a, int b) {
        Monoid L = M1, R = M1;
        for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
            if (a & 1) L = f(L, seg[a++]);
            if (b & 1) R = f(seg[--b], R);
        }
        return f(L, R);
    }

    Monoid operator[](const int& k) const {
        return seg[k + sz];
    }

    template< typename C >
    int find_subtree(int a, const C& check, Monoid& M, bool type) {
        while (a < sz) {
            Monoid nxt = type ? f(seg[2 * a + type], M) : f(M, seg[2 * a + type]);
            if (check(nxt)) a = 2 * a + type;
            else M = nxt, a = 2 * a + 1 - type;
        }
        return a - sz;
    }


    template< typename C >
    int find_first(int a, const C& check) {
        Monoid L = M1;
        if (a <= 0) {
            if (check(f(L, seg[1]))) return find_subtree(1, check, L, false);
            return -1;
        }
        int b = sz;
        for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
            if (a & 1) {
                Monoid nxt = f(L, seg[a]);
                if (check(nxt)) return find_subtree(a, check, L, false);
                L = nxt;
                ++a;
            }
        }
        return -1;
    }

    template< typename C >
    int find_last(int b, const C& check) {
        Monoid R = M1;
        if (b >= sz) {
            if (check(f(seg[1], R))) return find_subtree(1, check, R, true);
            return -1;
        }
        int a = sz;
        for (b += sz; a < b; a >>= 1, b >>= 1) {
            if (b & 1) {
                Monoid nxt = f(seg[--b], R);
                if (check(nxt)) return find_subtree(b, check, R, true);
                R = nxt;
            }
        }
        return -1;
    }
};
vel mul(vel& a, vel& b) {
    int n = a.size();
    int m = b.size();
    vel ans(n + m - 1, 0);
    rep(i, n) {
        rep(j, m) {
            ans[i + j] += a[i] * b[j];
            ans[i + j] %= p;
        }
    }
    return ans;
}
vel rui_p(vel& a, int n) {
    if (n == 0) { return { 1 }; }
    vel qans = rui_p(a, n / 2);
    qans = mul(qans, qans);
    if (n % 2 == 1) {
        qans = mul(qans, a);
    }
    return qans;
}
vvel disj_min(vel& v) {
    int n = v.size();
    vvel ret(22, vel(n));
    ret[0] = v;
    rep(i, 21) {
        rep(j, n) {
            int nex = j + (1 << i);
            if (nex < n) {
                ret[i + 1][j] = min(ret[i][j], ret[i][nex]);
            }
            else {
                ret[i + 1][j] = ret[i][j];
            }
        }
    }
    return ret;
}
vvel disj_max(vel& v) {
    int n = v.size();
    vvel ret(20, vel(n));
    ret[0] = v;
    rep(i, 19) {
        rep(j, n) {
            int nex = j + (1 << i);
            if (nex < n) {
                ret[i + 1][j] = max(ret[i][j], ret[i][nex]);
            }
            else {
                ret[i + 1][j] = ret[i][j];
            }
        }
    }
    return ret;
}
int find_min(vvel& dv, int l, int r) {
    int i = 19;
    while (l + (1 << i) > r) {
        i--;
    }
    return min(dv[i][l], dv[i][r - (1 << i)]);
}
int find_max(vvel& dv, int l, int r) {
    int i = 19;
    while (l + (1 << i) > r) {
        i--;
    }
    return max(dv[i][l], dv[i][r - (1 << i)]);
}
bool is_prime(int n) {
    if (n == 0 || n == 1) { return false; }
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) { return false; }
    }
    return true;
}
#define bs bitset<300>
void per(int& ans) {
    ans %= p;
    if (ans < 0) { ans += p; }
}
int count_zero(V<bs>& a) {
    int n = a.size();
    int now_st = 0;
    rep(i, 300) {

        int fir_ze = -1;
        for (int j = now_st; j < n; j++) {
            if (a[j][i]) { fir_ze = j; break; }
        }
        if (fir_ze >= 0) {
            rep(j, n) {
                if (j != fir_ze && a[j][i]) {
                    a[j] = a[fir_ze] xor a[j];
                }
            }
            swap(a[now_st], a[fir_ze]);
            now_st++;
        }
    }
    int ans = 0;
    rep(i, n) {
        bool fl = true;
        rep(j, 300) {
            if (a[i][j]) { fl = false; break; }
        }
        if (fl) { ans++; }
    }
    return ans;
}
void pri(vel& ans) {
    int n = ans.size();
    rep(i, n - 1) { cout << ans[i] << " "; }
    cout << ans[n - 1] << endl;
}
signed main() {
    int n; cin >> n;
    int now_min = 2000000;
    int ans = 0;
    rep(i, n) {
        int p; cin >> p;
        if (now_min >= p) { ans++; now_min = p; }
    }
    cout << ans << endl;
    return 0;
}