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=1000010,INF=0x7FFFFFFF;
int arr[SZ];

int turn(int x)
{
    if((x&1)==0)return x-1;
    else
    {
        return x+1;
    }
}

int main()
{
    std::ios::sync_with_stdio(0);
    //freopen("d:\\1.txt","r",stdin); 
    //for(;scanf("%d",&n)!=EOF;)
    {
        int n,m;
        cin>>n>>m;
        int one=1,two=2;
        for(int i=0;i<m;++i)
        {
            int type;
            cin>>type;
            if(type==1)
            {
                int step;
                cin>>step;
                one=(one+n+step-1)%n+1;
                two=(two+n+step-1)%n+1;
            }
            else
            {
                one=turn(one);
                two=turn(two);
            }
        }
        for(int i=0;i<n/2;++i)
        {
            arr[(one+2*i-1)%n+1]=1+2*i;
            arr[(two+2*i-1)%n+1]=2+2*i;
        }
        for(int i=1;i<=n;++i)
        {
            if(i!=1)cout<<" ";
            cout<<arr[i];
        }
        cout<<endl;
    }
    return 0;
}