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
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我们可以把模型抽象一下
有n-1个石子,一个人最多拿k个,问最后谁赢
——》裸的巴什博奕
#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN=1e6+10,INF=1e9+10;
int main()
{
#ifdef WIN32
freopen("a.in","r",stdin);
#else
#endif
int x,y;
while(scanf("%d%d",&x,&y))
{
if(x==0&&y==0) break;
if( (x-1)%(y+1)==0 ) printf("Jiang\n");
else printf("Tang\n");
}
return 0;
}