2017年ACM第八届山东省赛A题：Return of the Nim

Posted 2020-10-09 0一叶0知秋0

A题：Return of the Nim

时间限制: 1 秒  内存限制: 64 MB  |  提交: 33  解决: 16
 

 

题目描述

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

• There are n piles of stones denoted aspiles0,piles1,...,pilesn-1, 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

输入

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 piles₀,piles₁,...,piles_n-1.

• 1≤g≤15
• 2≤n≤30, where n is a prime.
• 1≤piles_i≤10⁵ where 0≤i≤n?1

输出

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

样例输入

2
3
2 3 2
2
2 1

样例输出

Sherlock
Watson
题意：有 T 组数据 
　　　每组数据 n 堆石子 
　　　每堆石子 num【i】
　　　sherlock先取  取到最后一颗石子为胜
　　　n == 2 为 威佐夫博弈
　　　n > 2 为 NIM博弈

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std ; 

#define maxn 100 
#define LL long long 
LL num[maxn] ; 

int main(){
    int n ; 
    int t ; 
    scanf("%d" , &t) ; 
    while(t--){
        scanf("%d" , &n) ; 
        for(int i=0 ; i<n ; i++){
            scanf("%d" , &num[i]) ; 
        }
        
        if(n==2){//威佐夫博弈
            if(num[0] < num[1]){
                swap(num[0],num[1]) ; 
            } 
            if((LL)((num[0]-num[1]) * (1.0 + sqrt(5.0))/2.0) == num[1]){
                printf("Watson\n") ; 
            }else printf("Sherlock\n") ; 
        }
        else {//NIM  游戏
            LL k = num[0] ; 
            for(int i=1 ; i<n ; i++){
                k = k ^ num[i] ; 
            } 
            if(k==0){
                printf("Watson\n") ; 
            }else printf("Sherlock\n") ; 
        }
        
    }
    return 0 ; 
}