链接：https://ac.nowcoder.com/acm/contest/5668/B
来源：牛客网
题目描述
Given a string S consists of lower case letters. You’re going to perform Q operations one by one. Each operation can be one of the following two types:
Modify: Given an integer x. You need to modify S according to the value of x. If x is positive, move the leftmost x letters in S to the right side of S; otherwise, move the rightmost |x| letters in S to the left side of S.
Answer: Given a positive integer x. Please answer what the x-th letter in the current string S is.
输入描述:
There are Q+2 lines in the input. The first line of the input contains the string S. The second line contains the integer Q. The following Q 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 c and an integer x. If c = ‘M’, this operation is a modify operation, that is, to rearrange S according to the value of x; if c = ‘A’, this operation is an answer operation, to answer what the x-th letter in the current string S is.
• 2 \le |S| \le 2 \times 10^62≤∣S∣≤2×10
6
(|S| stands for the length of the string S)
• S consists of lower case letters
• 1 \le Q \le 8 \times 10^51≤Q≤8×10
5
• c = 'M' or 'A'
• If c = 'M', 1 \le |x| < |S|1≤∣x∣<∣S∣
• If c = 'A', 1 \le x \le |S|1≤x≤∣S∣
• There is at least one operation in the input satisfies c = 'A'
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, S is ‘nowcoder’, six operations follow.
• The 1-st operation is asking what the 1-st letter is. The answer is 'n'.
• The 2-nd operation is to move the leftmost 4 letters to the rightmost side, so S is modified to 'odernowc'.
• The 3-rd operation is asking what the 6-th letter is. The answer is 'o'.
• The 4-th operation is to move the rightmost 3 letters to the leftmost side, so S is modified to 'owcodern'.
• The 5-th operation is to move the leftmost 1 letter to the rightmost side, so S is modified to 'wcoderno'.
• The 6-th operation is asking what the 1-st letter is. The answer is 'w
分析：
首先我们很容易使用图去模拟一下整个过程，
其实我们可以把字符串‘S’当成一个收尾相连的特殊字符串，其次我们考虑一下，怎样实现原数组不发生位置变化的同时使字符串S中的元素发生位置变化，说的有点扯哈，别急别急，我们都学过相对运动，假如我们把看成数组静止的，把其指向第一个元素的Num 进行移动，又因为字符串是一个环形，那么我们Num指向的可以看成初始的第一个元素，有它在我们便可以随时访问原数组了，不过需要一些计算，计算下面详细讲解。
紧接着，我们也看到了题意，输入分成了两个种，一种是A选项，主要是查第overset个元素是什么，打表。另一种是B选项，这个选项分为两种，一种是overset为负数，一种是为overset正数。
因此呢我们分开来考虑，
先考虑B选项：
```cpp
if(op[0]=='M'){
if(overset>0)
Num+=overset;
else{
Num+=(l+overset);
}
Num=Num%l;
}
```//注意：这里Num一直是指向数组的最初的首元素的。
再考虑A选项：
毕竟先A后再写B，必须将二者单独作为函数去实现，本人感觉麻烦，所以就想简便一点。
else {
long AP = (Num + overset) % l;
if(AP==0){
printf("%c\n" ,S[S.length()-1]);//这里你可以想一下AP=0
//意味着什么?也就是说指向原数组首地址的坐标值+移动的位数=数组的长度，~~是不是活到头了~~ ，
//所以是原数组最后一个元素了
}else{
printf("%c\n" ,S[(int)(AP-1)]);
}
}
总代码：
#include <bits/stdc++.h>
using namespace std;
int main(){
string S;
cin>>S;
int l=S.length();
int p,overset;
cin>>p;
long long Num=0;
char op[2];
for(int i=1;i<=p;i++){
scanf("%s%d",op,&overset);
if(op[0]=='M'){
if(overset>0)
Num+=overset;
else{
Num+=(l+overset);
}
Num=Num%l;
}
else {
long AP = (Num + overset) % l;
if(AP==0){
printf("%c\n" ,S[S.length()-1]);
}else{
printf("%c\n" ,S[(int)(AP-1)]);
}
}
}
}
欢迎指正！！！！！！！！