ati 時間復雜度 cond stdin idt span 題解 arch Language

題意

Language:DefaultGeorgia and Bob

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13237		Accepted: 4436

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

POJ Monthly--2004.07.18

分析

參照ACM_devil的題解。

我們把棋子按位置升序排列後，從後往前把他們兩兩綁定成一對。如果總個數是奇數，就把最前面一個和邊界（位置為0）綁定。 在同一對棋子中，如果對手移動前一個，你總能對後一個移動相同的步數，所以一對棋子的前一個和前一對棋子的後一個之間有多少個空位置對最終的結果是沒有影響的。於是我們只需要考慮同一對的兩個棋子之間有多少空位。我們把每一對兩顆棋子的距離(空位數)視作一堆石子，在對手移動每對兩顆棋子中靠右的那一顆時，移動幾位就相當於取幾個石子,與取石子遊戲對應上了,各堆的石子取盡,就相當再也不能移動棋子了。

我們可能還會考慮一種情況，就是某個玩家故意破壞，使得問題無法轉換為取石子，例如前一個人將某對中的前者左移，而當前玩家不將這對中的另一移動，則會導致本堆石子增多了，不符合nim。但是這種情況是不會出現的。因為贏家只要按照取石子進行即可獲勝，而輸家無法主動脫離取石子狀態。如果輸家想要讓某堆石子增多，那麽贏家只需要讓該堆減少回原狀，這樣輸家又要面臨跟上一回合同樣的局面。

時間復雜度\(O(n \log n)\)

代碼

#include<iostream>
#include<algorithm>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
    while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
co int N=1001;
int a[N];
void Georgia_and_Bob(){
    int n=read<int>(),x=0;
    for(int i=1;i<=n;++i) read(a[i]);
    std::sort(a+1,a+n+1);
    for(int i=n;i>0;i-=2) x^=a[i]-a[i-1]-1;
    puts(x?"Georgia will win":"Bob will win");
}
int main(){
//  freopen(".in","r",stdin),freopen(".out","w",stdout);
    for(int t=read<int>();t--;) Georgia_and_Bob();
    return 0;
}

POJ1704 Georgia and Bob