NBUT 1218 You are my brother

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

51 32 43 54 65 661 32 43 54 65 76 7

Sample Output

You are my elderYou are my brother

找祖先 看谁辈份大

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;int main(){    int la,lb,a,b,n,x,y,s1,s2;    while(scanf("%d",&n)!=EOF)    {        la=1;lb=2;s1=0;s2=0;        for(int i=0; i<n; i++)        {            scanf("%d%d",&x,&y);            if(la!=lb)            {                if(x==la)                {                    la=y;                    s1++;                }                if(x==lb)                {                    lb=y;                    s2++;                }            }        }        if(s1<s2)       printf("You are my younger\n");        else if(s1==s2) printf("You are my brother\n");        else if(s1>s2)  printf("You are my elder\n");    }    return 0;}