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

拿最後一個的人輸，anti-nim；

這題的關鍵在於可以把一堆得蘋果拿成兩堆的情況（使這堆不再連續），所以先手必勝的情況為：

①所有堆都為1且nim和為0；

②有一堆及以上不為1且nim和不為0；

對於第二個條件解釋如下：

①若只有一堆不為1，可以把這堆拿成一堆或兩堆1，並且nim和為1；
②若大於一堆不為1，可以操作把nim和變為0且還剩有知道一堆不為1。

ac程式碼：

#include <bits/stdc++.h>

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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