hdoj 2509 Be the Winner 【博弈】

Be the Winner Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2705    Accepted Submission(s): 1478 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

题意：n堆苹果，每次可以取任意一堆里任意数量的苹果，最后取完所有苹果的是失败者。双方都用最优策略，若先手赢输出Yes，反之No。

以前做过类似的题目，但还是WA了3次  o(╯□╰)o

注意：全为1的情况。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#define MAXN 32000+1
#define MAXM 100000+10
#define INF 0x3f3f3f3f
#define LL long long
#define debug printf("1\n");
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int main()
{
    int N;
    while(scanf("%d", &N) != EOF)
    {
        int num;
        int ans = 0;
        bool flag = false;
        for(int i = 0; i < N; i++)
        {
            scanf("%d", &num);
            ans ^= num;
            if(num != 1)
                flag = true;
        }
        if(!flag)//坑死 全为1。。。
        {
            if(N & 1)
                printf("No\n");
            else
                printf("Yes\n");
        }
        else
        {
            if(ans)
                printf("Yes\n");
            else
                printf("No\n");
        }
    }
    return 0;
}