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>
#include <stdlib.h>

typedef struct	s_p
{
  	int	n;
  	int	v;
}				t_p;

int	main(void)
{
  	int	N;
  	t_p	*p;
  	int	num;
  	int	i;
  	int j;
  	int	count;
  	int flg;
  
  	num = scanf("%d", &N);
  	if (num != 1)
   		return (0);
  
  	p = (t_p *)malloc(sizeof(t_p) * (N + 1));
  	for (i=1; i<=N; i++)
    {
      	num = scanf("%d", &p[i].n);
    	if (num != 1)
          	return (0);
      	p[i].v = 0;
    }
  	
  	flg = 0;
  	count = 1;
  	for (i=2; i<=N; i++)
    {
      	if (p[i - 1].n >= p[i].n)
        {
          	for (j=i - 1; j>0; j--)
            {
              	if (p[j].n >= p[i].n)
                {
                    flg = 1;
                    if (p[j].v == 1)
                    {
                        count++;
                        p[i].v = 1;
                        flg = 0;
                        break ;
                    }
                }
                else
                {
                    flg = 0;
                    break ;
                }
            }
            if (flg == 1)
            {
                count++;
                p[i].v = 1;
                flg = 0;
            }
        }
      	else 
        {
            ;
        }
    }
    printf("%d", count);
}