2015
09-17

# Stone

Tang and Jiang are good friends. To decide whose treat it is for dinner, they are playing a game. Specifically, Tang and Jiang will alternatively write numbers (integers) on a white board. Tang writes first, then Jiang, then again Tang, etc… Moreover, assuming that the number written in the previous round is X, the next person who plays should write a number Y such that 1 <= Y – X <= k. The person who writes a number no smaller than N first will lose the game. Note that in the first round, Tang can write a number only within range [1, k] (both inclusive). You can assume that Tang and Jiang will always be playing optimally, as they are both very smart students.

There are multiple test cases. For each test case, there will be one line of input having two integers N (0 < N <= 10^8) and k (0 < k <= 100). Input terminates when both N and k are zero.

There are multiple test cases. For each test case, there will be one line of input having two integers N (0 < N <= 10^8) and k (0 < k <= 100). Input terminates when both N and k are zero.

1 1
30 3
10 2
0 0

Jiang
Tang
Jiang

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

int main()
{
int n,k;
while (cin>>n>>k&&n+k)
{
n=n-1;
k=k+1;
n=n%k;
if (n!=0)
cout<<"Tang"<<endl;
else
cout<<"Jiang"<<endl;
}
}


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