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.
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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.
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.
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.
5
1 3
2 4
3 5
4 6
5 6
6
1 3
2 4
3 5
4 6
5 7
6 7
You are my elder You are my brother
#include<iostream> #include<cstring> using namespace std; int main() { int n; while(cin>>n) { int a,b,la=1,lb=2,da=0,db=0; for(int i=1;i<=n;i++) { cin>>a>>b; if(la==a){la=b;da++;} if(lb==a){lb=b;db++;} } if(db<da)cout<<"You are my elder"<<endl; else if(db==da) cout<<"You are my brother"<<endl; else if(db>da) cout<<"You are my younger"<<endl; } return 0; }
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原文地址：http://www.cnblogs.com/hyfer/p/5878078.html
