山东省第八届省赛 A：Return of the Nim（尼姆+威佐夫）

Problem 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:
1. Choose one pile and remove k(k >0) stones from it;
2. 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≤ 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

Sample Output

Sherlock
Watson

Hint

题意：

Sherlock 和 Watson 做游戏，游戏规则如下：

有n堆石子，第i堆石子的个数为pi， Sherlock 先取，有两种取法，

1.每次每人可取任意一石堆中的任意个数 (当然所取的个数不能超过石堆中石子的个数)

2.在没有出现空堆的情况下，每次每人可从所有石堆中取相同个数的石子 (当然所取的个数不能超过石堆中所含最少石子的个数)

谁先取完谁赢。告诉你石堆的个数及每堆所含石子的个数，要求你输出胜利者的姓名

题解：

两堆：威佐夫博弈（不作详细解释）

大于两堆：尼姆博弈

代码：

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <deque>
#include <stack>
#include <iomanip>
#include <cstdlib>
using namespace std;
#define is_lower(c) (c>='a' && c<='z')
#define is_upper(c) (c>='A' && c<='Z')
#define is_alpha(c) (is_lower(c) || is_upper(c))
#define is_digit(c) (c>='0' && c<='9')
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define IO ios::sync_with_stdio(0);\
    cin.tie();\
    cout.tie();
#define For(i,a,b) for(int i = a; i <= b; i++)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const ll inf=0x3f3f3f3f;
;
const ll inf_ll=(ll)1e18;
const ll maxn=100005LL;
const ll mod=1000000007LL;
+;
int ans[N];
int main()
{
    int T;
    cin >> T;
    while(T--)
    {
        int n;
        cin >> n;
        For(i, , n)
            cin >> ans[i];
        ){
            ] > ans[])
                swap(ans[], ans[]);
            ] - ans[]) * ((sqrt()+)/2.0));
            ])
                cout<< "Watson"<< endl;
            else
                cout<< "Sherlock" <<endl;
        } else {
            ;
            For(i, , n)
                res = res^ans[i];
            )
                cout<< "Watson" <<endl;
            else
                cout<< "Sherlock" <<endl;
        }
    }
    ;
}