Shuffle Cards

C: Shuffle Cards

时间限制: 1 Sec  内存限制: 128 MB
提交: 3  解决: 3
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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

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

Output one line contains N space-separated integers indicating the final order of the cards from top to bottom.

 

样例输入

5 1
2 3

 

样例输出

2 3 4 1 5

 使用rope（高效字符串处理数据结构）这种骚操作！

AC代码：

 1 #include<bits/stdc++.h>
 2 #include <ext/rope>
 3 using namespace std;
 4 using namespace __gnu_cxx;
 5 rope<int>ro;
 6 int main()
 7 {
 8     int n,m;
 9     scanf("%d %d",&n,&m);
10     for(int i=1; i<=n; i++)
11     {
12         ro.push_back(i);
13     }
14     int a,b;
15     while(m--)
16     {
17         scanf("%d %d",&a,&b);
18         ro=ro.substr(a-1,b)+ro.substr(0,a-1)+ro.substr(a+b-1,n-a-b+1);
19     }
20     for(int i=0; i<n-1; i++)
21     {
22         printf("%d ",ro[i]);
23     }
24     printf("%d\n",ro[n-1]);
25 }

View Code

 操作：
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

“rope的复制：

rope<char> *his[maxn];

his[0]=new rope<char>();

his[i]=new rope<char>(*his[i-1]);

his[i]->push_back(x);

”

转载于:https://www.cnblogs.com/lglh/p/9509994.html