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
题目大意就是 有 长度为n的数列，然后进行m次操作，输入x，y，将下标从x到y的子数列移动到开头；出m次操作后的数列
比赛的时候这个题可以用 rope卡过去，但比赛结束后，牛客加了数据超时了
先说下rope
声明
1）头文件
#include<ext/rope>
2)调用命名空间
using namespace __gnu_cxx;
rope的部分简单操作
| 函数 | 功能 |
| push_back(x) | 在末尾添加x |
| insert(pos,x) | 在pos插入x |
| erase(pos,x) | 从pos开始删除x个 |
| replace(pos,x) | 从pos开始换成x |
| substr(pos,x) | 提取pos开始x个 |
| at(x)/[x] | 访问第x个元素 |
小知识
先介绍几个可能使用到的函数
1）append()
string &append(const string &s,int pos,int n);//把字符串s中从pos开始的n个字符连接到当前字符串的结尾
或
a.append(b);
2）substr()
s.substr(0,5);//获得字符串s中从第零位开始长度为5的字符串（默认时长度为刚好开始位置到结尾）
定义/声明
rope<char> str;
<crope>r="abcdefg"
具体内容
总的来说，
1）运算符：rope支持operator += -= + - < ==
2）输入输出：可以用<<运算符由输入输出流读入或输出。
3）长度/大小：调用length()，size()都可以哦
友情提示:cena不支持rope
#include<iostream>
#include<ext/rope>
using namespace std;
using namespace __gnu_cxx;
rope<int>a;
int main(){
int n,m,x,y;
cin>>n>>m;
for(int i=1;i<=n;i++){
a.push_back(i);
}
for(int i=0;i<m;i++){
cin>>x>>y;
a.insert(0,a.substr(x-1,y));
a.erase(x+y-1,y);
}
for(int i=0;i<n;i++){
if(i==0){
cout<<a[i];
}else{
cout<<" "<<a[i];
}
}
cout<<endl;
}