牛客网暑期ACM多校训练营（第三场） C Shuffle Cards 平衡树 rope的运用

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

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 p

_i

, s

_i

 indicating that Eddy takes p

_i

-th card from top to (p

_i

+s

_i

-1)-th card from top(indexed from 1) and put them on the top of rest cards.


1 ≤ N, M ≤ 10

⁵

1 ≤ p

_i

 ≤ N
1 ≤ s

_i

 ≤ N-p

_i

+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


比赛的时候是模拟想到死，emmmmm，真的烦，一直没有做出来
比赛后看到别人用rope的代码
惊呆了
还是见识少了
这里补充下rope的知识

#include <ext/rope>
using namespace __gnu_cxx;
int a[1000];
rope<int> x;
rope<int> x(a,a + n);
rope<int> a(x);
 
x->at(10);
x[10];
x->push_back(x)     // 在末尾添加x
x->insert(pos,x)    // 在pos插入x
x->erase(pos,x)     // 从pos开始删除x个
x->replace(pos,x)   // 从pos开始换成x
x->substr(pos,x)    // 提取pos开始往后x个

　rope的原理是平衡树

　打算今晚做几个题目

贴一下AC代码

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iomanip>
#include <ext/rope>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
using namespace __gnu_cxx;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
rope<ll> r;
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    ll n, m;
    while( cin >> n >> m ) {
        r.clear();
        for( ll i = 1; i <= n; i ++ ) {
            r.push_back(i);
        }
        ll a, b;
        while( m -- ) {
            cin >> a >> b;
            r = r.substr(a-1,b) + r.substr(0,a-1) + r.substr(a+b-1,n-a-b+1);
        }
        for( ll i = 0; i < n-1; i ++ ) {
            cout << r[i] << " ";
        }
        cout << r[n-1] << endl;
    }
    return 0;
}

　　

转载于:https://www.cnblogs.com/l609929321/p/9374272.html