CodeForces 668B Little Artem and Dance

B. Little Artem and Dance

time limit per test

memory limit per test

input

output

Little Artem is fond of dancing. Most of all dances Artem likes rueda — Cuban dance that is danced by pairs of boys and girls forming a circle and dancing together.

n pairs of boys and girls standing in a circle. Initially, boy number 1 dances with a girl number 1, boy number 2dances with a girl number 2

1. xand some direction are announced, and all boys movex
2. 1 swaps with the one who was dancing with the girl number 2, while the one who was dancing with girl number 3swaps with the one who was dancing with the girl number 4 and so one. It's guaranteed thatn

Your task is to determine the final position of each boy.

Input

n and q (2 ≤ n ≤ 1 000 000, 1 ≤ q ≤ 2 000 000) — the number of couples in the rueda and the number of commands to perform, respectively. It's guaranteed that n

q lines contain the descriptions of the commands. Each command has type as the integer 1 or 2 first. Command of the first type is given as x ( - n ≤ x ≤ n), where 0 ≤ x ≤ n means all boys moves x girls in clockwise direction, while  - x means all boys move x

Output

n integers, the i-th of them should be equal to the index of boy the i-th girl is dancing with after performing all q

Examples

input

6 3
1 2
2
1 2

output

4 3 6 5 2 1

input

2 3
1 1
2
1 -2

output

1 2

input

4 2
2
1 3

output

1 4 3 2

先把1,2的位置确定了，

别的数字的位置变动和1,2肯定是一样的

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
#define MAX 1000000
int n,q;
int c[MAX+5];
int x,y;
int main()
{
    scanf("%d%d",&n,&q);
    int a=1;int b=2;
    for(int i=1;i<=q;i++)
    {
       scanf("%d",&x);
       if(x==1)
       {
           scanf("%d",&y);
           a=(a+y+n)%n;
           if(a==0)
               a=n;
           b=(b+y+n)%n;
           if(b==0)
               b=n;
       }
       else
       {
           if(a&1) a+=1;
           else a-=1;
           if(b&1) b+=1;
           else  b-=1;
       }
    }
    a-=1;
    b-=2;
    for(int i=1;i<=n;i++)
    {
        if(i&1)
        {
            int pos=(i+a)%n;
            if(pos==0)
                pos=n;
            c[pos]=i;
        }
        else
        {
            int pos=(i+b)%n;
            if(pos==0)
                pos=n;
            c[pos]=i;
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(i==n)
            printf("%d\n",c[i]);
        else
            printf("%d ",c[i]);
    }
    return 0;

}