POJ 1704中，如何运用Georgia and Bob-Nim博弈策略？
- 内容介绍
- 文章标签
- 相关推荐
本文共计954个文字，预计阅读时间需要4分钟。
Georgia和Bob决定玩一个自创的游戏。他们在纸上画了一排网格，从左到右依次编号为1、2、3、……，并在不同的网格上放置N个棋子，如图所示。
Georgia and Bob
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
231 2 381 5 6 7 9 12 14 17
Sample Output
Bob will winGeorgia will win
Source
POJ Monthly--2004.07.18
Time Limit:1000MS
Memory Limit:10000K
Total Submissions:6341
Accepted:1826
先将数一一配对（a,b)，若是奇数则把最左边与0配对。
那么若对方向左挪a，我们也挪动b相同步数。
那么唯一会影响的就是各配对间的间隔。
假设有间隔c1,c2..cn
那么相当于每次选一个数，减去至少1.
这就像nim取石，有n堆石子，每次至少取走其中一堆的至少一个，谁最先没石子取就输。
这个问题的答案是 c1^c2^..^cn=0时有必输解，否则必胜。
都懂得……
证明：
假设你面对一个c1^c2^...^cn=0的局面
A:取走若干石子使c1^c2^...^cn=k
B:取走ci的最高位与k的最高位相同的那堆 ci^k个 //最高位的1^1=0,所以ci^k<ci
A:恭喜你还是必输
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
int T,n,a[100000+10];
int main()
{
scanf("%d",&T);
while (T--)
{
scanf("%d",&n);
For(i,n) scanf("%d",&a[i]);
sort(a+1,a+1+n);
int tmp=0,head=1;
if (n%2) tmp=a[1]-1,head++;
for(int i=head;i<=n;i+=2)
{
tmp^=(a[i+1]-a[i]-1);
}
if (tmp) puts("Georgia will win");
else puts("Bob will win");
}
return 0;
}
本文共计954个文字，预计阅读时间需要4分钟。
Georgia和Bob决定玩一个自创的游戏。他们在纸上画了一排网格，从左到右依次编号为1、2、3、……，并在不同的网格上放置N个棋子，如图所示。
Georgia and Bob
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
231 2 381 5 6 7 9 12 14 17
Sample Output
Bob will winGeorgia will win
Source
POJ Monthly--2004.07.18
Time Limit:1000MS
Memory Limit:10000K
Total Submissions:6341
Accepted:1826
先将数一一配对（a,b)，若是奇数则把最左边与0配对。
那么若对方向左挪a，我们也挪动b相同步数。
那么唯一会影响的就是各配对间的间隔。
假设有间隔c1,c2..cn
那么相当于每次选一个数，减去至少1.
这就像nim取石，有n堆石子，每次至少取走其中一堆的至少一个，谁最先没石子取就输。
这个问题的答案是 c1^c2^..^cn=0时有必输解，否则必胜。
都懂得……
证明：
假设你面对一个c1^c2^...^cn=0的局面
A:取走若干石子使c1^c2^...^cn=k
B:取走ci的最高位与k的最高位相同的那堆 ci^k个 //最高位的1^1=0,所以ci^k<ci
A:恭喜你还是必输
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
int T,n,a[100000+10];
int main()
{
scanf("%d",&T);
while (T--)
{
scanf("%d",&n);
For(i,n) scanf("%d",&a[i]);
sort(a+1,a+1+n);
int tmp=0,head=1;
if (n%2) tmp=a[1]-1,head++;
for(int i=head;i<=n;i+=2)
{
tmp^=(a[i+1]-a[i]-1);
}
if (tmp) puts("Georgia will win");
else puts("Bob will win");
}
return 0;
}