AtCoder Beginner Contest 273(E)

合集 - acm(93)

1.AtCoder Beginner Contest 293(C,D ,E,F)2023-03-172.Educational Codeforces Round 115 (Rated for Div. 2)（D,E)2023-03-183.AtCoder Beginner Contest 294(E,F,G)2023-04-034.AtCoder Beginner Contest 2462023-03-295.中国石油大学（北京）第三届“骏码杯”程序设计竞赛（同步赛）(D,E,F)2023-03-256.Codeforces Global Round 16(D,E,F)2023-03-257.AtCoder Beginner Contest 209（D，E）2023-05-088.Monoxer Programming Contest 2022（AtCoder Beginner Contest 238）（E，F）2023-05-079.AtCoder Beginner Contest 285（B,D,E,F)2023-05-0610.AtCoder Beginner Contest 242（D，E)2023-05-0311.AtCoder Beginner Contest 223（D,E,F)2023-04-1512.AtCoder Beginner Contest 207（D,E)2023-04-1113.AtCoder Beginner Contest 247（E,F)2023-04-1014.AtCoder Beginner Contest 226(E,F,G)2023-04-0515.AtCoder Beginner Contest 229(F,G)2023-06-23

16.AtCoder Beginner Contest 273(E)2023-06-13

17.AtCoder Beginner Contest 286(G)2023-06-0218.AtCoder Beginner Contest 287(C,D,E,F)2023-06-0119.AtCoder Beginner Contest 288(D,E,F)2023-05-3120.AtCoder Beginner Contest 289(E,F)2023-05-3021.AtCoder Beginner Contest 290(D,E)2023-05-2922.AtCoder Beginner Contest 292(E,F,G)2023-05-2823.AtCoder Beginner Contest 298(D,F)2023-05-2724.AtCoder Beginner Contest 299(E,F)2023-05-2725.AtCoder Beginner Contest 300(E,F)2023-05-2526.AtCoder Beginner Contest 302（E，F，G）2023-05-2427.AtCoder Beginner Contest 253(E,F)2023-05-1728.AtCoder Beginner Contest 245(D，E，F)2023-05-1529.AtCoder Beginner Contest 248(D，E，F) 2023-05-1430.AtCoder Beginner Contest 206（Sponsored by Panasonic）（E，F）2023-05-0931.Codeforces Round 875 (Div. 2)(D)2023-07-0832.AtCoder Beginner Contest 178(E,F)2023-07-0833.AtCoder Beginner Contest 307（E,F,G)2023-07-0134.CodeTON Round 5 (Div. 1 + Div. 2, Rated, Prizes!)C2023-06-3035.Educational Codeforces Round 151 (Rated for Div. 2)(C,D)2023-06-3036.AtCoder Beginner Contest 212(E,F)2023-06-2437.牛客小白月赛512022-12-0938.2022年浙大城市学院新生程序设计竞赛（同步赛）（补题）2022-12-1139.Codeforces Round #770 (Div. 2)B，C2022-12-2440.Codeforces Global Round 24(B,C)2022-12-2341.Codeforces Round #836 (Div. 2）C2022-12-2142.Codeforces Round #840 (Div. 2) C2022-12-2043.Good Bye 2022: 2023 is NEAR C2022-12-3144.Codeforces Round #765 (Div. 2)A，B，C2022-12-3145.Codeforces Round #766 (Div. 2)C，D2022-12-2946.Codeforces Round #841 (Div. 2) and Divide by Zero 20222022-12-2847.Codeforces Round #767 (Div. 2)C ,D 2022-12-2748.Codeforces Round #768 (Div. 2)C ,D2022-12-2649. Codeforces Round #769 (Div. 2) B，C2022-12-2550.Educational Codeforces Round 122 (Rated for Div. 2)，C，D2022-12-2551.Educational Codeforces Round 119 (Rated for Div. 2)2023-01-1452.AtCoder Beginner Contest 2582023-01-1353.Codeforces Round #763 (Div. 2)C2023-01-1254.Codeforces Round #843 (Div. 2)（B，C，D，E）2023-01-1255.Educational Codeforces Round 141 (Rated for Div. 2)（B，C，D）2023-01-1056.AtCoder Beginner Contest 275（B，C，D，E，F）2023-01-0957.Codeforces Round #842 (Div. 2)（B，D，E）2023-01-0958.AtCoder Beginner Contest 284（D，E，F）2023-01-0859.The 14th Jilin Provincial Collegiate Programming Contest（补题）2023-01-0760.牛客小白月赛65（C，D，E，F）2023-01-0761.AtCoder Beginner Contest 281（D，E，F）2023-01-0562.Good Bye 2021: 2022 is NEAR D2023-01-0563.Hello 20232023-01-0464.The 15th Jilin Provincial Collegiate Programming Contest(补题)2023-01-0365.Codeforces Round #781 (Div. 2)C 2023-01-0266.Hello 2022（B，D）2023-01-0167.AtCoder Beginner Contest 272（D,E)2023-03-1068.Codeforces Round 751 (Div. 2)（D）2023-03-0869.Codeforces Round 856 (Div. 2)(C,D)2023-03-0870.Codeforces Round 752 (Div. 2)（C，D，E）2023-03-0871.Codeforces Round 855 (Div. 3)（E，F）2023-03-0572.AtCoder Regular Contest 131（A,B,C）2023-03-0573.Educational Codeforces Round 144 (Rated for Div. 2)（A，B，C，D）2023-03-0574.Codeforces Round 853 (Div. 2)（C,D)2023-03-0375.牛客练习赛109（C,D)2023-03-0376.AtCoder Beginner Contest 291（Sponsored by TOYOTA SYSTEMS）（D，E，F）2023-03-0177.Educational Codeforces Round 143 (Rated for Div. 2)（A，C,D)2023-02-2878.Codeforces Round #852 (Div. 2)（C,D)2023-02-1379.Educational Codeforces Round 118 (Rated for Div. 2)（D，E）2023-02-0980.AtCoder Beginner Contest 236（D，E，F）2023-02-0981.Codeforces Round #850 (Div. 2, based on VK Cup 2022 - Final Round)(B,D)2023-02-0882.Codeforces Round #848 (Div. 2)（B,C,D）2023-02-0783.TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) (B,C,D) 2023-02-0784.Codeforces Round #846 (Div. 2)（B，E） 2023-02-0785.Educational Codeforces Round 142 (Rated for Div. 2)（C，D）2023-02-0486.2023牛客寒假算法基础集训营62023-02-0487.Codeforces Round #845 (Div. 2) and ByteRace 2023（A，B，C）2023-02-0288.2023牛客寒假算法基础集训营5 2023-02-0289.2023牛客寒假算法基础集训营3 2023-01-2190.2023牛客寒假算法基础集训营22023-01-1991.2023牛客寒假算法基础集训营12023-01-1892.Educational Codeforces Round 120 (Rated for Div. 2) C，D2023-01-1593.AtCoder Beginner Contest 254（C，D，E，F) 2023-01-15

收起

AtCoder Beginner Contest 273(E)

E(链式结构，思维)

E

题目大意就是原本有一个空的序列，我们后面会进行$q$次操作，每次操作我们都需要输出此时的序列的最后数字

下面有几种操作

$ADD$ $x$,代表在这在这个序列的最后面添加一个$x$

$DELETE$,代表如果此时的序列存在数字的话，那么我们就移除最后一个数字

$SAVE$ $y$,把此时的序列记录在一个题目中给出的一个$pag{e}_y$

$LOAD$ $z$:把此时的序列变成在$pag{e}_z$里面保存的序列里面

我之前是想过用$map<int,vector<int>>$，但是它$re$了

我$ac$的解法用的是链式结构

我们先用一个$now$随着每次操作的标记

添加一个数字，我们会在数组里面添加上$x$，而且我们还会保存此时的$now$标记，这样我们后面在实现保存$page$和获取$page$里面的数组，我们每次保存就是保存$now$,后面假如我们需要用到用到$page$,我们就会调用里面的标记$now$,然后把这个值赋给此时的$now$，同样删除一个数字就是把当前的$now$赋给它的上一个$now$

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include<cmath>
#include <unordered_map>
#include <array>
#include <cstring>
#include <bitset>
#include <numeric>
using namespace std;
#define int long long 
#define LL long long
#define ios  ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define inf 1e18
#define INF 1e18
#define eps 1e-6
#define mem(a,b) memset((a),(b),sizeof(a))
const int maxn=5e5+10;
const int mod=998244353;
int q;
map<int,int>bk;
vector<pair<int,int>>res;
int now;
signed main ()
{
  ios;
  cin>>q;
  res.push_back({0,-1});
  now=0;
  while (q--)
  {
    string op;
    cin>>op;
    if(op=="ADD")
    {
      int x;
      cin>>x;
      res.push_back({now,x});
      now=res.size()-1;
    }
    else if(op=="DELETE")
    {
      now=res[now].first;
    }
    else if(op=="SAVE")
    {
      int x;
      cin>>x;
      bk[x]=now;
    }
    else if(op=="LOAD")
    {
      int x;
      cin>>x;
      now=bk[x];
    }
    cout<<res[now].second<<" ";
  }
  system ("pause");
  return 0;
}