[poj2975]Nim

Nim

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2835		Accepted: 1232

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 removingone 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 arek1 +k2 + … +
kn possible moves. We write eachki in binary (base 2). Then, the Nim position is losing if and only if, among all theki’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 thexor of theki’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:

111
1011
1101

There are an odd number of 1’s among the rightmost digits, so this position is not losing. However, supposek3 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 whenk1 = 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 withn = 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

Source
Stanford Local 2005

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题意:
题意跟这道题一样:http://blog.csdn.net/hccz95/archive/2011/03/29/6286266.aspx
只是求的不是简单的是否有必胜策略,而是要求先手第一步有几种情况是必胜策略.
只要求出
sum=a[1] xor a[2] xor...xor a
,
若sum=0,则先手必输;
若sum>0,先手可以进行操作是sum = 0,到达一个先手必输态,操作如下:
设a[i]' = sum xor a[i], 若a[i]' < a[i], 那么可以把a[i]变成a[i], 此时sum'=sum xor a[i] xor a[i]'=0







{Source Code

Problem: 2975 User: hccz95
Memory: 848K Time: 16MS
Language: Pascal Result: Accepted

Source Code
}
var n , ans : Longint ;
a : array[ 0..1000 ] of Longint ;

procedure main;
var t , i , x : Longint ;
begin
while true do
begin
readln( n ); if n = 0 then break;
x := 0 ; ans := 0 ;
for i := 1 to n do
begin
read( a[ i ] );
x := x xor a[ i ] ;
end;
if x > 0 then
for i := 1 to n do if x xor a[ i ] < a[ i ] then
inc( ans );
writeln( ans );
end;
end;

begin
main;
end.