Be the Winner HDU - 2509

最新推荐文章于 2022-12-03 19:48:18 发布

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最新推荐文章于 2022-12-03 19:48:18 发布

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本文探讨了在特定苹果分组游戏中赢得比赛的策略。玩家轮流取走成行的苹果，目标是避免取走最后一颗。通过分析苹果堆的数量与大小，文章提供了判断先手是否能必胜的方法。

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Be the Winner
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4387 Accepted Submission(s): 2436

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

和上一篇基本一样，代码基本没改。。。。。
题目： John HDU - 1907

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#define ll long long

using namespace std;



const int MAXN=1000000;


int main(){
    int g, c, n, T;
    while(scanf("%d",&n) != EOF) {
     int a, sum = 0;
     g = c = 0;
     for(int i = 0; i < n; i++)
     {
         scanf("%d",&a);
         sum ^= a;
         if(a == 1) g ++;
         else c ++;
     }
     if((sum == 0&& c >= 2) || ((sum != 0)&&(c == 0)&&(g %2 != 0))) 
         puts("No");
     else 
         puts("Yes");
    }
    return 0;
}