HDU5090——贪心——Game with Pearls

[align=left]Problem Description[/align]
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”.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each game, output a line containing either “Tom” or “Jerry”.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

Jerry
Tom

[align=left]Source[/align]
2014上海全国邀请赛——题目重现（感谢上海大学提供题目）

大意：给出n个管子，以及管子里面珍珠的数目，问你向里面假如0或者k的倍数，是否能够满足i = a[i]，如果都满足输出Jerry，没有考虑其他也可以转变成这个并不是一一对应的关系

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[150];
int b[150];
int main()
{
int T;
int n,k;
scanf("%d",&T);
while(T--){
memset(b,0,sizeof(b));
scanf("%d%d",&n,&k);
for(int i = 1; i <= n ; i++)
scanf("%d",&a[i]);
for(int i = 1; i <= n ; i++){
for(int p = a[i]; p <= n ; p+=k){
b[p]++;
}
}
int flag = 1;
for(int i = 1; i <= n && flag ; i++){
if(!b[i]){
flag = 0;
break;
}
else {
for(int j = i; j <= n ; j+=k)
b[j] --;
}
}
if(flag)
printf("Jerry\n");
else printf("Tom\n");
}
return 0;
}