Problem Description
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
Source
题意：给你n个数，(1<=n<=100),还有一个k(1<=k<=n)，然后允许给某一个数加上k的正整数倍，当然可以不加，问你是否可以把这n个数变成1,2,3,...,n,可以就输出Jerry,否则输出Tom。
分析：一开始直接开个100数组去记录每个数出现的次数。。。如果某一个数出现了，那就不用动这个数了，对于没出现的数，枚举要加的数(k,2k,3k,...注意边界，不能超过这个数)。如果存在某一个数没出现而且不能通过别的数加若干个k,直接判断不能实现 。。简简单单就过了
来源：http://mathlover.info/archives/hdu5090
代码：
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 using namespace std;
5
6 int main()
7 {
8 int m,n,k;
9 cin>>m;
10 int a[105],cnt[102];
11 bool p[105];
12 while(m--)
13 {
14 cin>>n>>k;
15 memset(cnt,0,sizeof(cnt));
16 memset(p,0,sizeof(p));
17
18 for(int i=0;i<n;i++)
19 {
20 cin>>a[i];
21 if(p[a[i]]==0)///判断出现的标志
22 p[a[i]]=1;
23 else
24 cnt[a[i]]++;///多余的数字
25 }
26
27 bool flag=1;///标记谁能赢
28 for(int i=1;i<=n;i++)
29 {
30 if(p[i]==0)///如果这个数没有出现
31 {
32 bool ok=0;
33 for(int j=1;i-j*k>=1;j++)///依次减去k的倍数
34 {
35 if(cnt[i-j*k])///若多余
36 {
37 cnt[i-j*k]--;
38 ok=1;
39 break;///即是可以由多余的数字变化而来，逆过程
40 }
41 }
42 if(!ok)///ok==0-------若果没有找到，则进行此if
43 {
44 flag=0;
45 break;
46 }
47 }
48 }
49
50 if(flag)
51 puts("Jerry");
52 else
53 puts("Tom");
54 }
55 return 0;
56 }