poj 2975 Nim

Nim

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5284 Accepted: 2470

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, …, k_n stones respectively; in such a position, there are k₁ + k₂ + … + k_n possible moves. We write each k_i in binary (base 2). Then, the Nim position is losing if and only if, among all the k_i’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 xorof the k_i’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:

 11110111101 

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 ≤ k_i ≤ 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

37 11 1321000000000 10000000000

Sample Output

30

Source

Stanford Local 2005

题解：具体分析过程参见nim游戏的证明。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int a[11000],n,k;int main(){  scanf("%d",&n);  while (n!=0)   {    memset(a,0,sizeof(a));    for (int i=1;i<=n;i++)     scanf("%d",&a[i]);    k=a[1];    for (int i=2;i<=n;i++)     k^=a[i];    int ans=0;    for (int i=1;i<=n;i++)     if ((a[i]^k)<a[i])      ans++;    printf("%d\n",ans);    scanf("%d",&n);   }}