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Description
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 2otherwise.
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：依次给你i个数，问到最后一个人留下的数组中，所有的数字都没办法分解成两个数就算输。
解法：数字分解成两个数，不管你怎么分，它的分解次数是相同的，那么先手要胜利，只能是在他那次留下奇数个分解次数
2的分解次数为1,3的为2,4的为3，n的分解次数为n-1；
我们依次加起来，判断奇偶就行
#include<bits/stdc++.h>
using namespace std;
long long a[100005];
int n,m;
long long sum[1000005];
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
sum[i]=sum[i-1]+(a[i]-1);
}
for(int i=1;i<=n;i++)
{
if(sum[i]%2==1)
{
cout<<"1"<<endl;
}
else
{
cout<<"2"<<endl;
}
}
return 0;
}
Codeforces Round #366 (Div. 2) B
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原文地址：http://www.cnblogs.com/yinghualuowu/p/5751478.html
