2020牛客暑期多校B

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≤∣S∣≤2×106times 10^6 (|S| stands for the length of the string S)
    •  S consists of lower case letters
    •   1≤Q≤8×105times 
    •  c = 'M' or 'A'
    •  If c = 'M', 1≤∣x∣<∣S∣
    •  If c = 'A', 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

复制

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

输出

复制

n
o
w

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'. 代码：

思路：其实很简单他相当于一个字符串的位置不断变化而且是连续的所以我们只要取模就可以了

#include <bits/stdc++.h>

using namespace std;
#define int long long

void solve()
{
    string s;
    cin >> s;
    int n;
    cin >> n;
    int res = 0,pos = s.size();
    for(int i = 1;i <= n;i ++)
    {
        string op;
        cin >> op;
        int x;
        cin  >> x;
        if(op == "A")
        {
            cout << s[(res+x-1)%pos] << '\n';
        }
        else
        {
            res += x;
            res = (res%pos+pos)%pos;
        }
    }
}
signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int _ = 1;
    while(_--)
    solve();
}