Nim(POJ_2975)

Description

Nim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s move consists of removing one or more stones from any single pile. Play ends when all the stones have been removed, at which point the last player to have moved is declared the winner. Given a position in Nim, your task is to determine how many winning moves there are in that position.

A position in Nim is called “losing” if the first player to move from that position would lose if both sides played perfectly. A “winning move,” then, is a move that leaves the game in a losing position. There is a famous theorem that classifies all losing positions. Suppose a Nim position contains n piles having k1, k2, …, kn stones respectively; in such a position, there are k1 + k2 + … + kn possible moves. We write each ki in binary (base 2). Then, the Nim position is losing if and only if, among all the ki’s, there are an even number of 1’s in each digit position. In other words, the Nim position is losing if and only if the xor of the ki’s is 0.

Consider the position with three piles given by k1 = 7, k2 = 11, and k3 = 13. In binary, these values are as follows:

0111
1011
1101

There are an odd number of 1’s among the rightmost digits, so this position is not losing. However, suppose k3 were changed to be 12. Then, there would be exactly two 1’s in each digit position, and thus, the Nim position would become losing. Since a winning move is any move that leaves the game in a losing position, it follows that removing one stone from the third pile is a winning move when k1 = 7, k2 = 11, and k3 = 13. In fact, there are exactly three winning moves from this position: namely removing one stone from any of the three piles.

Input

The input test file will contain multiple test cases, each of which begins with a line indicating the number of piles, 1 ≤ n ≤ 1000. On the next line, there are n positive integers, 1 ≤ ki ≤ 1, 000, 000, 000, indicating the number of stones in each pile. The end-of-file is marked by a test case with n = 0 and should not be processed.

Output

For each test case, write a single line with an integer indicating the number of winning moves from the given Nim position.

Sample Input

3
7 11 13
2
1000000000 1000000000
0

Sample Output

3
0

一开始没看懂题意，后来看懂了却没有什么思路，研究别人的题解也没研究出什么来，他们讲的都太深奥，看不懂，最后终于有一份题解点醒了我。题目大意是让求在当前状态下有多少种方法能让先手进入必胜态。我们知道当两堆石子数量相同时肯定是后手必胜，所以如果想让先手进入必胜态就是当其中一堆石子数量大于另一堆时，先手从数量多的那一堆中取出一些石子让两堆石子数量相同，这样先手必胜。而我们现在面临的是N堆石子，可以这样，把这N堆石子分成a[i]和a[1] + … + a[i - 1] + a[i + 1] + … + a[n]两堆，是winning moves的唯一条件就是a[i] > a[1] ^ … ^ a[i - 1] ^ a[i + 1] ^ … ^ a[n]，然后从a[i]中取出一些石子让两者相等，所以到最后就是求有多少个a[i]满足这个条件。

代码

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int main(){    int n;    int a[1100];    while(scanf("%d", &n) && n)    {        int ans = 0;        for(int i = 0; i < n; i++)        {            scanf("%d", &a[i]);            ans ^= a[i];        }        int cnt = 0;        for(int i = 0; i < n; i++)        {            if((a[i] ^ ans) < a[i]) //a[i] ^ ans等于什么不用说了吧？这个都想不出来的话回去重学异或吧= =                cnt++;        }        printf("%d\n", cnt);    }    return 0;}