Stone
Total Submission(s): 109 Accepted Submission(s): 85
題意:
兩個人寫數字,要求當前個寫的數字Y與前一個寫的數字X滿足:思路:
1. 1 <= Y - X <= k
2. Y要求小於N(第一次取只需滿足1<=Y<=k);
不都滿足則輸。
可將題目意思理解爲取石子(題目有提示),取的石子是有規定的,誰能取到第N-1個,那麼就能贏得比賽。
對了,就是巴什博弈。
/*************************************************************************
> File Name: Stone.cpp
> Author: BSlin
> Mail:
> Created Time: 2013年09月28日 星期六 12時22分50秒
************************************************************************/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <iterator>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#define MP make_pair
#define INF (1<<30)
#define PI acos(-1.0)
#define esp 1e-8
const int dx[4]={0,0,0,0};
using namespace std;
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)
#define LL __int64
#define LLS "%" "I" "6" "4" "d"
#else
#define LL long long
#define LLS "%" "l" "l" "d"
#endif
int main(int argc, char** argv) {
//read;
int n,k,ans;
while(scanf("%d%d",&n,&k)){
if(n == 0 && k == 0) break;
if(n == 1) {
printf("Jiang\n");
continue;
}
else {
ans = (n - 1) % (k + 1);
if(ans == 0) {
printf("Jiang\n");
}
else printf("Tang\n");
}
}
return 0;
}