HDU 2509 Be the Winner（Anti-Nim）

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

拿最后一个的人输，anti-nim；

这题的关键在于可以把一堆得苹果拿成两堆的情况（使这堆不再连续），所以先手必胜的情况为：

①所有堆都为1且nim和为0；

②有一堆及以上不为1且nim和不为0；

对于第二个条件解释如下：

①若只有一堆不为1，可以把这堆拿成一堆或两堆1，并且nim和为1；
②若大于一堆不为1，可以操作把nim和变为0且还剩有知道一堆不为1。

ac代码：

#include 

using namespace std;

#define rep(i,a,n) for(int i = (a); i < (n); i++)
#define per(i,a,n) for(int i = (n)-1; i >= (a); i--)
#define clr(arr,val) memset(arr, val, sizeof(arr))
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define pi acos(-1)
typedef pair<int, int> pii;
typedef long long LL;
const LL mod = 1000000007;
const double eps = 1e-8;

int main(int argc, char const *argv[]) {
    int n, tmp;
    while (cin >> n) {
        bool f = true;
        int ans = 0;
        rep(i, 0, n){
            scanf("%d", &tmp);
            ans ^= tmp;
            if(tmp > 1) f = false;
        }
        if(f) puts(ans?"NO":"Yes");
        else puts(ans?"Yes":"No");
    }
    return 0;
}

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