codechef Fun with Rotation

Problem Description

You are given an array A of N integers. You are to fulfill M queries. Each query has one of the following three types:

• C d : Rotate the array A clockwise by d units.
• A d : Rotate the array A anticlockwise by d units.
• R d : Query for the value of the element, currently being the d-th in the array A.

Input

The first line contains two numbers - N and M respectively.

The next line contains N space separated Integers, denoting the array A.

Each of the following M lines contains a query in the one of the forms described above.

Output

For each query of type R output the answer on a separate line.

Constraints

• 1 ≤ N ≤ 100000
• 1 ≤ M ≤ 100000
• 1 ≤ d ≤ N, in all the queries
• 1 ≤ elements of A ≤ 1000000
• The array A and the queries of the type R are 1-based.

 

Example

Input:5 55 4 3 3 9R 1C 4R 5A 3R 2Output:533

Explanation

The initial array : 5 4 3 3 9

The answer for R 1 : 5

The array after C 4 : 9 5 4 3 3

The answer for R 5 : 3

The array after A 3 : 4 3 3 9 5

The answer for R 2 : 3 

题解

话说这道题还蛮容易让人想到链表的，看这数据范围，链表很虚啊。

有时候就是这么简单。

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<cmath>#include<algorithm>using namespace std;int m,n,a[100002];char ch[3];void init(){scanf("%d%d",&n,&m);int i;for(i=1;i<=n;i++) scanf("%d",&a[i]);}void work(){int i,lr=0,rr=0,x,st=1,w;for(i=1;i<=m;i++)   {scanf("%s%d",ch,&x);    if(ch[0]=='C') lr+=x;    else if(ch[0]=='A') rr+=x;    else        {if(lr>rr)       {w=(lr-rr)%n;    st=(st+w)%n;    if(st==0) st=n;   }else   {w=(rr-lr)%n;    st=(st+n-w)%n;    if(st==0) st=n;   }w=(st+x-1)%n;if(w==0) w=n;printf("%d\n",a[w]);lr=rr=0;   }   }}int main(){init(); work();return 0;}