hdu2509 Be the Winner

Be the Winner

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1543    Accepted Submission(s): 817

Problem Description

Let's consider m apples divided into n groups. Each group contains no more than 100 apples, arranged in a line. You can take any number of consecutive apples at one time.
For example "@@@" can be turned into "@@" or "@" or "@ @"(two piles). two people get apples one after another and the one who takes the last is 
the loser. Fra wants to know in which situations he can win by playing strategies (that is, no matter what action the rival takes, fra will win).

 

Input

You will be given several cases. Each test case begins with a single number n (1 <= n <= 100), followed by a line with n numbers, the number of apples in each pile. There is a blank line between cases.

 

Output

If a winning strategies can be found, print a single line with "Yes", otherwise print "No".

 

Sample Input

   
   
   
   

    
    
    
    2
2 2
1
3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    No
Yes
   
   
   
   

 

Source

ECJTU 2008 Autumn Contest

 

Recommend

lcy

和上一题 是一样的!

首先给出结论：先手胜当且仅当 ①所有堆石子数都为1且游戏的SG值为0（即有偶数个孤单堆-每堆只有1个石子数）；②存在某堆石子数大于1且游戏的SG值不为0. 
证明： 
   
   
   
   

    
    
    
    
若所有堆都为1且SG值为0，则共有偶数堆石子，故先手胜。 
    
    
    
    
 
     
 
      i)只有一堆石子数大于1时，我们总可以对该石子操作，使操作后堆数为奇数且所有堆的石子数均为1；  
     
ii)有超过一堆的石子数1时，先手将SG值变为0即可，且总还存在某堆石子数大于1 
   
   
   
   
因为先手胜。 

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define M 50
int pri[M];
int main()
{
   int tcase,i,ans,re,n;

   while(scanf("%d",&n)!=EOF){

        ans=0,re=0;
        for(i=0;i<n;i++){
            scanf("%d",&pri[i]);
            if(pri[i]>1)
            ans++;
            re^=pri[i];
        }
        if(ans){
            if(re)
            printf("Yes\n");
            else
            printf("No\n");
        }
        else {
            if(re)
            printf("No\n");
            else
            printf("Yes\n");
        }
   }
    return 0;
}