HDU 4764 Stone(巴什博奕)

2018-02-22 20:09:26来源:cnblogs.com作者:自为风月马前卒人点击

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2026    Accepted Submission(s): 1428


Problem DescriptionTang 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. InputThere 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. OutputFor each case, print the winner's name in a single line. Sample Input1 130 310 20 0 Sample OutputJiangTangJiang Source2013 ACM/ICPC Asia Regional Changchun Online  Recommendliuyiding   |   We have carefully selected several similar problems for you:  6263 6262 6261 6260 6259   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;}

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