HDU

Tom and Jerry are playing a game with tubes and pearls. The rule of the game is: 

1) Tom and Jerry come up together with a number K. 

2) Tom provides N tubes. Within each tube, there are several pearls. The number of pearls in each tube is at least 1 and at most N. 

3) Jerry puts some more pearls into each tube. The number of pearls put into each tube has to be either 0 or a positive multiple of K. After that Jerry organizes these tubes in the order that the first tube has exact one pearl, the 2nd tube has exact 2 pearls, …, the Nth tube has exact N pearls. 

4) If Jerry succeeds, he wins the game, otherwise Tom wins. 

Write a program to determine who wins the game according to a given N, K and initial number of pearls in each tube. If Tom wins the game, output “Tom”, otherwise, output “Jerry”.

Input

The first line contains an integer M (M<=500), then M games follow. For each game, the first line contains 2 integers, N and K (1 <= N <= 100, 1 <= K <= N), and the second line contains N integers presenting the number of pearls in each tube.

Output

For each game, output a line containing either “Tom” or “Jerry”.

Sample Input

2 5 1 1 2 3 4 5 6 2 1 2 3 4 5 5 

Sample Output

Jerry Tom 

题意是：给出n个数和一个整数K，那n个数每个只能加上k的正倍数或者不加，问能否使序列变成1~n的序列；

两种方法，都是一步一步往前推，不放过任何情况；

第一种：与桶排序，用桶计数，一步一步往前推；

#include<stdio.h>      #include<string.h>#include<algorithm>using namespace std;          //像这样的题就是一步一步的推得，不然情况太多了 int main(){int i,j,k,t,n,tt;scanf("%d",&t);int vis[110];while(t--){memset(vis,0,sizeof(vis)); // 用数组，与桶排序相似，桶计数； scanf("%d%d",&n,&k);for(i=0;i<n;i++){scanf("%d",&tt);vis[tt]++;}int flag=1;   for(i=1;i<=n;i++){if(!vis[i]){flag = 0;break;}vis[i+k]+=(vis[i]-1); // -1 是只允许用一个，多的数一步一步往前推，不放过任何情况； }if(flag) printf("Jerry\n");else printf("Tom\n");}return 0;}

第二种加的是a[i],标记a[i],也是一步一步往上加的

#include<stdio.h>      #include<string.h>   #include<algorithm>using namespace std;int main()      {int i,j,k,t,n;scanf("%d",&t);int vis[210],a[110];while(t--){scanf("%d%d",&n,&k);  for(i=0;i<n;i++)scanf("%d",&a[i]);memset(vis,0,sizeof(vis));for(i=0;i<n;i++)  //这个也是一步一步往上加，不放过任何情况,不过这次是用数组标记，加的是a[i]的值； {while(1){if(a[i]>n) break;if(!vis[a[i]]){vis[a[i]]=1;break;}a[i]+=k;}}sort(a,a+n);for(i=0;i<n;i++){if(a[i]!=i+1)break;}if(i>=n) printf("Jerry\n");else printf("Tom\n");}}