hdu 2509 Nim变形

Be the Winner

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

 

有n堆苹果，每次可以从一堆中取任何数量的苹果，最先拿完的输。

Nim问题是求最先拿完的赢，但这个变形了，最先拿完的输。被称之为anti-nim博弈论。

证明就不说了，直接说结论。

当所有的数都是1的话，奇异局势不为0就赢，当有大于1的数时，奇异局势为0才赢。

#include <bits/stdc++.h>
using namespace std;

int main() {
    int m;
    while(cin >> m) {
        int ans = 0, x, flag = 0;
        while(m--) {
            cin >> x;
            ans ^= x;
            if(x > 1) flag = 1;
        }
        if((ans&&flag)||(!ans&&!flag)) printf("Yes
");
        else printf("No
");
    }
    return 0;
}