Time Limit: 1000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
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
2014上海全国邀请赛――题目重现(感谢上海大学提供题目)
分析:贪心即可,升序排序然后扫一遍,如果当前数等于下标就检查下一个,比下标大则Tom赢,比下标小就加上k然后重复上述过程。所有数都扫过了就Jerry赢。也有二分图匹配的做法。
AC代码:
#include <cstdio> #include <algorithm> using namespace std; int n, k, a[110]; void input() { scanf("%d%d", &n, &k); for (int i = 1; i <= n; i++) scanf("%d", &a[i]); } bool cmp(const int x, const int y) { return x < y; } void solve() { bool flag = 1, ok = 0; while (!ok) { sort(a + 1, a + 1 + n, cmp); for (int i = 1; i <= n; i++) { if (a[i] > i) { flag = 0; ok = 1; break; } if (a[i] != i) { a[i] += k; break; } if (i == n) ok = 1; } } if (flag) printf("Jerry\n"); else printf("Tom\n"); } int main() { int T; scanf("%d", &T); while (T--) { input(); solve(); } return 0; }
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