2018牛客暑期多校训练第三场——C Shuffle Cards（rope大法）

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld

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 ≤ 105
1 ≤ pi ≤ N
1 ≤ 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.

 

示例1

输入

5 1
2 3

输出

2 3 4 1 5

示例2

输入

5 2
2 3
2 3

输出

3 4 1 2 5

示例3

输入

5 3
2 3
1 4
2 4

输出

3 4 1 5 2

rope相关操作详细介绍：Rope大法好

小结：

有M个操作，每个操作都是从p开始的s个数移动到开头，问你最后的序列是什么。

很容易看出区间切割问题。找一个Splay伸展树的板子应该也能过。这也是这种题目的裸题了。不过赛后听说了一项黑科技 Rope这个东西也是STL里的在C++11下可以使用。具体用法可以看下面这几个博客。 
参考博客1 参考博客2 
真的是黑科技啊。对于这个题目简直是秒杀，直接每次都把对应区间进行拼接就可以了。复杂度是M*sqrt(N) 

代码：

#include 
#include 
 
using namespace std;
using namespace __gnu_cxx;
 
rope R; 
 
int main(){
	
	int N,M;
	while(scanf("%d %d",&N,&M) == 2){
		R.clear();
		for(int i=1 ; i<=N ; ++i)R.push_back(i);
		int a,b;
		for(int i=1 ; i<=M ; ++i){
			scanf("%d %d",&a,&b);
			if(a == 1)continue;
			else R = R.substr(a-1,b) + R.substr(0,a-1) + R.substr(a-1+b,N-(a-1+b));
		}
		for(int i=0 ; i