牛客网暑期ACM多校训练营（第三场）  C   Shuffle Cards

题目：

链接：https://www.nowcoder.com/acm/contest/141/C
来源：牛客网

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
复制
5 1
2 3
输出
复制
2 3 4 1 5
示例2
输入
复制
5 2
2 3
2 3
输出
复制
3 4 1 2 5
示例3
输入
复制
5 3
2 3
1 4
2 4
输出
复制
3 4 1 5 2

思路：

　　rope可当做可持久化平衡树，适用于大量、冗长的串操作

　基本操作：

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

　　4）插入/添加等：

　　push_back(x);//在末尾添加x

　　insert(pos,x);//在pos插入x，自然支持整个char数组的一次插入

　　erase(pos,x);//从pos开始删除x个

　　copy(pos,len,x);//从pos开始到pos+len为止用x代替

　　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

代码：

#include<cstdio>
#include<ext/rope>  //固定写法
using namespace std;
using namespace __gnu_cxx;  //固定写法
rope<int> ss;  //实质是可持久化平衡树int n,m;
int main()
{scanf("%d%d",&n,&m);for(int i=1; i<=n; i++) ss.push_back(i);  //放入元素（1~n）while(m--){int p,s;scanf("%d%d",&p,&s);ss = ss.substr(p-1,s)+ss.substr(0,p-1)+ss.substr(p+s-1,n-p-s+1); //重新组合三个区间 substr(起始位置，区间长度)
    }for(int i=0; i<n; i++){printf("%d",ss[i]);if(i==n-1)puts("");else printf(" ");}return 0;
}