Stone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 545 Accepted Submission(s): 405


Problem Description 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.

Input 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.

Output For each case, print the winner's name in a single line.

Sample Input

1 1
30 3
10 2
0 0

Sample Output

Jiang
Tang
Jiang

Source 2013 ACM/ICPC Asia Regional Changchun Online

題意:tang和jiang比賽，tang先取一個1 - k的數字，之後每人取一個數字要求比上一個數字多1-k。先取到》=n的人輸掉，問獲勝者。

思路:推理，就是搶30的游戲，可以控制每兩個人增長（1 + k)，所以只要判斷(n - 1) % (1 + k) == 0，就可以判斷輸贏。

代碼:

#include 
#include 

int main() {
    int n, k;
    while (~scanf("%d%d", &n, &k) && n || k) {
	if ((n - 1) % (k + 1) == 0)
	    printf("Jiang\n");
	else printf("Tang\n");
    }
}