spy——B - You are my brother

Description
Little A gets to know a new friend, Little B, recently. One day, they realize that they are family 500 years ago. Now, Little A wants to know whether Little B is his elder, younger or brother.
Input
There are multiple test cases.
For each test case, the first line has a single integer, n (n<=1000). The next n lines have two integers a and b (1<=a,b<=2000) each, indicating b is the father of a. One person has exactly one father, of course. Little A is numbered 1 and Little B is numbered 2.
Proceed to the end of file.
Output

For each test case, if Little B is Little A’s younger, print “You are my younger”. Otherwise, if Little B is Little A’s elder, print “You are my elder”. Otherwise, print “You are my brother”. The output for each test case occupied exactly one line.

Sample Input
5
1 3
2 4
3 5
4 6
5 6
6
1 3
2 4
3 5
4 6
5 7

6 7

Sample Output
You are my elder

You are my brother

 <iostream> #include  <cstring> #include  <algorithm> #include  <cstdio> using  namespace std ; int  main () { int n ; int ans [ 2005 ]; while ( scanf ( "%d" ,& n )!= EOF ) { memset ( ans , 0 , sizeof ( ans )); for ( int i = 0 ; i < n ; i ++) {   int a , b ;   scanf ( "%d%d" ,&a a ,& b );   ans [ a ]= b ; } int temp = 1 ; int num1 = 0 ; int num2 = 0 ; while ( ans [ temp ]!= 0 ) { num1 ++; temp = ans [ temp ]; } temp = 2 ; while ( ans [ temp ]!= 0 ) { num2 ++; temp = ans [ temp ]; } if ( num1 > num2 ) printf ( "You是我的长辈\ n “ ）; 否则 ，如果（ NUM1 == NUM2 ）的printf （“你是我的兄弟\ n “ ）; 否则    的printf （“你是我年轻的\ n “ ）; } 返回 0 ; }