c++中STL的rope的使用

Rope其主要是结合了链表和数组各自的优点，链表中的节点指向每个数据

块，即数组，并且记录数据的个数，然后分块查找和插入。在g++头文件中，< ext / rope >中有成型的块状链表，在using namespace

__gnu_cxx;空间中，其操作十分方便。

　　基本操作：
rope test;

test.push_back(x);//在末尾添加x

test.insert(pos,x);//在pos插入x　　

test.erase(pos,x);//从pos开始删除x个

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

test.replace(pos,x);//从pos开始换成x

test.substr(pos,x);//提取pos开始x个

test.at(x)/[x];//访问第x个元素

其算法复杂度n*(n^0.5)，可以在很短的时间内实现快速的插入、删除和查找字符串，是一个很厉害的神器！

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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

可用rope代替块状链表求解，大大提高了速度

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<ext/rope>
using namespace std;
using namespace __gnu_cxx;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int N = 100005;

int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        rope<int> str;
        for(int i = 0;i < n;++i)
        {
            str.push_back(i + 1);
        }
        for(int i = 0;i < m;++i)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            str = str.substr(a - 1,b) + str.substr(0,a - 1) + str.substr(a + b - 1,n - a - b + 1);
        }
        for(int i = 0;i < n;++i){
            if(i == n - 1){
                printf("%d\n",str[i]);
            }else{
                printf("%d ",str[i]);
            }
        }
    }
    return 0;
}