Georgia and Bob
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 7233 | Accepted: 2173 |
Description
Georgia and Bob decide to play a self-invented
game. They draw a row of grids on paper, number the grids from left to right by
1, 2, 3, ..., and place N chessmen on different grids, as shown in the following
figure for example:
Georgia
and Bob move the chessmen in turn. Every time a player will choose a chessman,
and move it to the left without going over any other chessmen or across the left
edge. The player can freely choose number of steps the chessman moves, with the
constraint that the chessman must be moved at least ONE step and one grid can at
most contains ONE single chessman. The player who cannot make a move loses the
game.
Georgia always plays first since "Lady first". Suppose that
Georgia and Bob both do their best in the game, i.e., if one of them knows a way
to win the game, he or she will be able to carry it out.
Given the
initial positions of the n chessmen, can you predict who will finally win the
game?
Input
The first line of the input contains a single
integer T (1 <= T <= 20), the number of test cases. Then T cases follow.
Each test case contains two lines. The first line consists of one integer N (1
<= N <= 1000), indicating the number of chessmen. The second line contains
N different integers P1, P2 ... Pn (1 <= Pi <= 10000), which are the
initial positions of the n chessmen.
Output
For each test case, prints a single line, "Georgia
will win", if Georgia will win the game; "Bob will win", if Bob will win the
game; otherwise ‘Not sure‘.
Sample Input
2
3
1 2 3
8
1 5 6 7 9 12 14 17
Sample Output
Bob will win
Georgia will win
Source
转化成nim游戏，两个石子之间的间隔就是石子个数
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5
6 using namespace std;
7
8 const int MAX_N = 1005;
9 int p[MAX_N];
10
11 int main()
12 {
13 int t;
14 scanf("%d",&t);
15 while(t--) {
16 int N;
17 scanf("%d",&N);
18 for(int i = 0; i < N; ++i) {
19 scanf("%d",&p[i]);
20 }
21 if(N % 2 == 1) p[N++] = 0;
22 sort(p, p + N);
23
24 int x = 0;
25 for(int i = 0; i + 1 < N; i += 2) {
26 x ^= (p[i + 1] - p[i] - 1);
27 }
28
29 if(x == 0) printf("Bob will win\n");
30 else printf("Georgia will win\n");
31 }
32 //cout << "Hello world!" << endl;
33 return 0;
34 }