2020牛客暑期多校训练营（第三场）B

题目描述

  Given a string consists of lower case letters. You're going to perform operations one by one. Each operation can be one of the following two types.
  Modify: Given an integer , you need to modify according to the value of . If is positive, move the leftmost letters in to the right side of ; otherwise, move the rightmost letters in to the left side of .
  Answer: Given a positive integer . Please answer what the -th letter in the current string is.

输入描述

  There are lines in the input. The first line of the input contains the string . The second line contains the integer . The following lines each denotes an operation. You need to follow the order in the input when performing those operations.
  Each operation in the input is represented by a character and an integer . If c=='M', this operation is a modify operation, that is, to rearrange according to the value of ; if c=='A', this operation is an answer operation, to answer what the -th letter in the current string is.
   ( stands for the length of the string ).
   consists of lower case letters.
  .
   or . If , . If , .
  There is at least one operation in the input satisfies .

输出描述

  For each answer operation, please output a letter in a separate line representing the answer to the operation. The order of the output should match the order of the operations in the input.

示例1

输入

nowcoder
6
A 1
M 4
A 6
M -3
M 1
A 1

输出

n
o
w

备注

  Initially, is "nowcoder", six operations follow.
  The -st operation is asking what the -st letter is. The answer is 'n'.
  The -nd operation is to move the leftmost letters to the rightmost side, so is modified to "odernowc".
  The -rd operation is asking what the -th letter is. The answer is 'o'.
  The -th operation is to move the rightmost letters to the leftmost side, so is modified to "owcodern".
  The -th operation is to move the leftmost letter to the rightmost side, so is modified to "wcoderno".
  The -th operation is asking what the -st letter is. The answer is 'w'.

分析

  将字符串看作一个环，定义一个指针 指向字符串的起点，从起点开始逆时针遍历整个环，得到的就是当前字符串；不论如何更改，环的结构是不会变的。查询时，从起点开始，逆时针经过 个点，最终到达的点即为询问的答案；更新时，若 ，要将字符串末尾的某个字符作为起点，将指针顺时针移动 次；否则，将指针逆时针移动 次。环上的移动，通过对字符串长度 取模来体现。

代码

/******************************************************************
Copyright: 11D_Beyonder All Rights Reserved
Author: 11D_Beyonder
Problem ID: 2020牛客暑期多校训练营（第三场） Problem B
Date: 8/14/2020
Description: Pointer Simulation
*******************************************************************/
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=2000005;
int _;
char s[N];
int q;
int len;
int main(){
    scanf("%s%d",s,&q);
    len=strlen(s);
    int pointer=0;
    while(q--){
        char c;
        int x;
        getchar();
        scanf("%c%d",&c,&x);
        if(c=='M'){
            //取模体现环上移动的性质
            pointer=(pointer+x+len)%len;
        }else{
            printf("%c\n",s[(pointer+x-1)%len]);
        }
    }
    return 0;
}