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题目链接：CF 705B
题面：
Peter Parker wants to play a game with Dr. Octopus. The game is about cycles. Cycle is a sequence of vertices, such that first one is connected with the second, second is connected with third and so on, while the last one is connected with the first one again. Cycle may consist of a single isolated vertex.
Initially there are k cycles, i-th of them consisting of exactly vi vertices. Players play alternatively. Peter goes first. On each turn a player must choose a cycle with at least 2 vertices (for example, x vertices) among all available cycles and replace it by two cycles with p and x?-?p vertices where 1?≤?p?<?x is chosen by the player. The player who cannot make a move loses the game (and his life!).
Peter wants to test some configurations of initial cycle sets before he actually plays with Dr. Octopus. Initially he has an empty set. In the i-th test he adds a cycle with ai vertices to the set (this is actually a multiset because it can contain two or more identical cycles). After each test, Peter wants to know that if the players begin the game with the current set of cycles, who wins?
Peter is pretty good at math, but now he asks you to help.
The first line of the input contains a single integer n (1?≤?n?≤?100?000) — the number of tests Peter is about to make.
The second line contains n space separated integers a1,?a2,?...,?an (1?≤?ai?≤?109), i-th of them stands for the number of vertices in the cycle added before the i-th test.
Print the result of all tests in order they are performed. Print 1 if the player who moves first wins or 2 otherwise.
3 1 2 3
2 1 1
5 1 1 5 1 1
2 2 2 2 2
In the first sample test:
In Peter‘s first test, there‘s only one cycle with 1 vertex. First player cannot make a move and loses.
In his second test, there‘s one cycle with 1 vertex and one with 2. No one can make a move on the cycle with 1 vertex. First player can replace the second cycle with two cycles of 1 vertex and second player can‘t make any move and loses.
In his third test, cycles have 1, 2 and 3 vertices. Like last test, no one can make a move on the first cycle. First player can replace the third cycle with one cycle with size 1 and one with size 2. Now cycles have 1, 1, 2, 2 vertices. Second player‘s only move is to replace a cycle of size 2 with 2 cycles of size 1. And cycles are 1, 1, 1, 1, 2. First player replaces the last cycle with 2 cycles with size 1 and wins.
In the second sample test:
Having cycles of size 1 is like not having them (because no one can make a move on them).
In Peter‘s third test: There a cycle of size 5 (others don‘t matter). First player has two options: replace it with cycles of sizes 1 and 4 or 2 and 3.
So, either way first player loses.
题意：
博弈游戏，一堆数，可以拆成两堆非空的，所有堆都不能拆，则为输。每次不断加入一个新堆，问在当前所有堆的情况下的输赢状况。
解题：
因为一个数字n，可以拆的次数为固定的n-1次，故而只要统计全部的可操作次数，看总和的奇偶性即可。
代码：
#include <iostream> #include <cstdio> #define LL long long using namespace std; int main() { int n; LL sum=0,tmp; cin>>n; while(n--) { cin>>tmp; sum+=tmp-1; if(sum%2) cout<<1<<endl; else cout<<2<<endl; } return 0; }
【打CF，学算法——二星级】Codeforces 705B Spider Man (简单博弈)
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原文地址：http://blog.csdn.net/david_jett/article/details/52148981
