Problem:
Score : 200 points

Problem Statement

There are N rabbits, numbered 1 through N.

The i-th (1≤i≤N) rabbit likes rabbit a_i.
Note that no rabbit can like itself, that is, a_i≠i.

For a pair of rabbits i and j (i＜j), we call the pair (i，j) a friendly pair if the following condition is met.

Rabbit i likes rabbit j and rabbit j likes rabbit i.

Calculate the number of the friendly pairs.

Constraints

2≤N≤10^5

1≤a_i≤N

a_i≠i

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the number of the friendly pairs.

Sample Input 1

4
2 1 4 3

Sample Output 1

2

There are two friendly pairs: (1，2) and (3，4).

Sample Input 2

3
2 3 1

Sample Output 2

0

There are no friendly pairs.

Sample Input 3

5
5 5 5 5 1

Sample Output 3

1

There is one friendly pair: (1，5).

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)


#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, const char * argv[]) {
   
    
    int N;
    scanf("%d\n",&N);
    
    int i;
    
    int *a;
   
    a=(int *)malloc(N+1);
    
    int aa=1;

    char *buff;
     buff=(char *)malloc(256);
    
    char *line;
   
    
    if ( fgets(buff,sizeof(buff),stdin) != NULL )
    {//←端末からの入力
        
        for ( line = strtok(buff," ") ; line != NULL ; line = strtok(NULL," ") )
        {/*
               　　分割された文字列 line が得られる
               　　*/
            
           // printf("%s\n",line);
            a[aa]=atoi(line);
          // printf("a[%d]cast %d\n",aa,a[aa]);
            aa++;
            
            
            }
    }
    
   // free(buff);
    
    
    
    
    int likesum=0;
    
    for (i=1; i<=N; i++) {
        
       // printf("  %d    %d \n",a[i],a[a[i]]);
        
            if (i==a[a[i]]) {
                likesum++;
                //printf("\na");
            }
            
        
    }
    

    printf("%d\n",likesum/2);


//    free(buff);
//    free(a);
    
    return 0;
}