省SD2017 A Return of the Nim【威佐夫博弈+NIM】

Return of the Nim

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 9 Solved: 3

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Description

Sherlock and Watson are playing the following modified version of Nim game:

There are n piles of stones denoted as , and n is a prime number;

Sherlock always plays first, and Watson and he move in alternating turns. During each turn, the current player must perform either of the following two kinds of moves:

Choose one pile and remove k(k >0) stones from it;

Remove k stones from all piles, where 1≤k≤the size of the smallest pile. This move becomes unavailable if any pile is empty.

Each player moves optimally, meaning they will not make a move that causes them to lose if there are still any better or winning moves.

Giving the initial situation of each game, you are required to figure out who will be the winner

Input

The first contains an integer, g, denoting the number of games. The 2×g subsequent lines describe each game over two lines:

1. The first line contains a prime integer, n, denoting the number of piles.

2. The second line contains n space-separated integers describing the respective values of , .

1≤g≤15

2≤n≤30, where n is a prime.

1≤pilesi≤ 105 where 0≤i≤n−1

Output

For each game, print the name of the winner on a new line (i.e., either “Sherlock” or “Watson”)

Sample Input

2

3

2 3 2

2

2 1

Sample Output

Sherlock

Watson

HINT

Source

山东省第八届ACM程序设计大赛2017.5.7

这题，优秀！

//威佐夫博弈+NIM

给你素数堆石子，你每次可以在一个堆中取任意数量石子，也可以在所有堆中去掉相同数量的石子。

如果只有两个堆，不必说，直接就是威佐夫博弈了。

然后这个素数堆是关键，除去2，剩下的堆不仅是素数也是奇数。

NIM

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int a[1000];
int main()
{
int t;
cin >> t;
while (t--)
{
int n;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> a[i];
}
if (n == 2)//威佐夫博弈
{
if (a[0] < a[1])
{
swap(a[0], a[1]);
}
if (floor((a[0] - a[1])*((sqrt(5.0) + 1.0) / 2.0)) != a[1])
{
cout << "Sherlock" << endl;

}
else
cout << "Watson" << endl;
}
else//NIM
{
int k = a[0];
for (int i = 1; i < n; i++)
{
k ^= a[i];
}
if (k == 0)
cout << "Watson" << endl;
else
cout << "Sherlock" << endl;
}
}
return 0;
}