??？途W暑期ACM多校訓練營（第三場）? C? ?Shuffle Cards

題目：

鏈接：https://www.nowcoder.com/acm/contest/141/C
來源：牛客網

時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 262144K，其他語言524288K
Special Judge, 64bit IO Format: %lld
題目描述
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].
輸入描述:
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
輸出描述:
Output one line contains N space-separated integers indicating the final order of the cards from top to bottom.
示例1
輸入
復制
5 1
2 3
輸出
復制
2 3 4 1 5
示例2
輸入
復制
5 2
2 3
2 3
輸出
復制
3 4 1 2 5
示例3
輸入
復制
5 3
2 3
1 4
2 4
輸出
復制
3 4 1 5 2

思路：

　　rope可當做可持久化平衡樹，適用于大量、冗長的串操作

　基本操作：

? ? ? ?1）運算符：rope支持operator += -= + - < ==

　　2）輸入輸出：可以用<<運算符由輸入輸出流讀入或輸出。

　　3）長度/大小：調用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，自然支持整個char數組的一次插入

　　erase(pos,x);//從pos開始刪除x個

　　copy(pos,len,x);//從pos開始到pos+len為止用x代替

　　replace(pos,x);//從pos開始換成x

　　substr(pos,x);//提取pos開始x個

　　at(x)/[x];//訪問第x個元素

代碼：

#include<cstdio> #include<ext/rope> //固定寫法 using namespace std; using namespace __gnu_cxx; //固定寫法 rope<int> ss; //實質是可持久化平衡樹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); //重新組合三個區間 substr(起始位置，區間長度) }for(int i=0; i<n; i++){printf("%d",ss[i]);if(i==n-1)puts("");else printf(" ");}return 0; }

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轉載于:https://www.cnblogs.com/longl/p/9381126.html

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