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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#define _CRT_SECURE_NO_WARNINGS

int main(void){

    int n, i;
    int a[200000];
    printf("\n");
    //scanf("%d", &n);
    
    //n값을 입력받고,
    //printf를 n번 반복하는 함수
    //이 떄, a 값은 최대치로 설정해둔다.
    
    if(scanf("%d", &n)==1)
    {
    for(i=1; i<n+1; i++)
    {
        printf("\n");
        scanf("%d", &a[i]);//a[1], a[2], ... a[n]의 값을 읽어들인다

    }
    }
    //박스 i는 박스 j보다 크면 안 된다.
    
    int sum= n;
    
    for(i=1; i<n+1; i++)//i가 1, 2, 3,...,n일 때 작업 반복
    {
        
        for(int j=1; j<i+1; j++)
        {
            
            if(a[i]>a[j]){
                //모든 a[i]<=a[j]==1
                sum= sum-1;
                break;
            }
            else{
                sum= sum;
            }
        }
    }
    
    printf("%d", sum);
        
    return 0;
}