codeforces668b //Little Artem and Dance// Codeforces Round #348

题意：2种操作，转动或者奇偶位互换。

不论怎么交换，1的后两位一定是3,3的后两位一定是5。因此只要记录1,2的位置。

//#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include <stack>
using namespace std;
const int SZ=,INF=0x7FFFFFFF;
int arr[SZ];

int turn(int x)
{
    if((x&)==)return x-;
    else
    {
        return x+;
    }
}

int main()
{
    std::ios::sync_with_stdio();
    //freopen("d:\\1.txt","r",stdin);
    //for(;scanf("%d",&n)!=EOF;)
    {
        int n,m;
        cin>>n>>m;
        int one=,two=;
        for(int i=;i<m;++i)
        {
            int type;
            cin>>type;
            if(type==)
            {
                int step;
                cin>>step;
                one=(one+n+step-)%n+;
                two=(two+n+step-)%n+;
            }
            else
            {
                one=turn(one);
                two=turn(two);
            }
        }
        for(int i=;i<n/;++i)
        {
            arr[(one+*i-)%n+]=+*i;
            arr[(two+*i-)%n+]=+*i;
        }
        for(int i=;i<=n;++i)
        {
            if(i!=)cout<<" ";
            cout<<arr[i];
        }
        cout<<endl;
    }
    return ;
}