题目链接：https://www.nowcoder.com/acm/contest/141/C
题目描述
Eddy likes to play cards game since there are always lots of randomness in the game. For most of the cards game, the very first step in the game is shuffling the cards. And, mostly the randomness in the game is from this step. However, Eddy doubts that if the shuffling is not done well, the order of the cards is predictable!
To prove that, Eddy wants to shuffle cards and tries to predict the final order of the cards. Actually, Eddy knows only one way to shuffle cards that is taking some middle consecutive cards and put them on the top of rest. When shuffling cards, Eddy just keeps repeating this procedure. After several rounds, Eddy has lost the track of the order of cards and believes that the assumption he made is wrong. As Eddy‘s friend, you are watching him doing such foolish thing and easily memorizes all the moves he done. Now, you are going to tell Eddy the final order of cards as a magic to surprise him.
Eddy has showed you at first that the cards are number from 1 to N from top to bottom.
For example, there are 5 cards and Eddy has done 1 shuffling. He takes out 2-nd card from top to 4-th card from top(indexed from 1) and put them on the top of rest cards. Then, the final order of cards from top will be [2,3,4,1,5].
输入描述:
The first line contains two space-separated integer N, M indicating the number of cards and the number of shuffling Eddy has done.Each of following M lines contains two space-separated integer pi, si indicating that Eddy takes pi-th card from top to (pi+si-1)-th card from top(indexed from 1) and put them on the top of rest cards. 1 ≤ N, M ≤ 1051 ≤ pi ≤ N1 ≤ si ≤ N-pi+1
输出描述:
Output one line contains N space-separated integers indicating the final order of the cards from top to bottom.
case_1
Input:
5 1
2 3
Output:
2 3 4 1 5
case_2
Input:
5 2
2 3
2 3
Output:
3 4 1 2 5
题目大意：
给一串长度为N的连续上升序列， K次操作，每次把开头为 st， 长度为 len 的子串与前面的调换。
求经过K次调换，最后的这串东西是什么。
官方题解：二叉平衡树
Main Idea: Data structure, Implementation
We need to keep cutting out some consecutive number and put them in front of the rest list.
Since we need to quick find (p_i)-th element, it would be too slow with linked-list.
The solution is just maintaining the list with treap(or any other balanced binary tree).
Overall Time complexity: O((N+M) \lg N) Overall Space complexity: O(N)
大概思路：
①STL神器 —— rope 时间上勉强解，实现简单粗暴。
②splay （未完待续）
①AC code（4824K 859MS）：
1 #include<bits/stdc++.h>
2 #include<ext/rope>
3 using namespace std;
4 using namespace __gnu_cxx;
5 rope<int>A;
6 int main (void){
7 int n,m;
8 scanf("%d %d",&n,&m);
9 for(int i=1;i<=n;i++){
10 A.push_back(i);
11 }
12 while(m--){
13 int l,r;
14 scanf("%d %d",&l,&r);
15 l--;
16 A=A.substr(l,r)+A.substr(0,l)+A.substr(l+r,n-l-r);
17 }
18 for(int i=0;i<n;i++)
19 printf("%d ",A[i]);
20 }
原文地址：https://www.cnblogs.com/ymzjj/p/9393752.html