靠这把上了蓝

A. Palindromic Supersequence
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string A. Find a string B, where B is a palindrome and A is a subsequence of B.

A subsequence of a string is a string that can be derived from it by deleting some (not necessarily consecutive) characters without changing the order of the remaining characters. For example, "cotst" is a subsequence of "contest".

A palindrome is a string that reads the same forward or backward.

The length of string B should be at most 104. It is guaranteed that there always exists such string.

You do not need to find the shortest answer, the only restriction is that the length of string B should not exceed 104.

Input

First line contains a string A (1 ≤ |A| ≤ 103) consisting of lowercase Latin letters, where |A| is a length of A.

Output

Output single line containing B consisting of only lowercase Latin letters. You do not need to find the shortest answer, the only restriction is that the length of string B should not exceed 104. If there are many possible B, print any of them.

Examples
input

Copy
aba
output
aba
input

Copy
ab
output
aabaa
Note

In the first example, "aba" is a subsequence of "aba" which is a palindrome.

In the second example, "ab" is a subsequence of "aabaa" which is a palindrome.

输出一个字符串是输入串的子串,并且是回文串,不要求最短

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+;
int main()
{
ios::sync_with_stdio(false);
string s;
cin>>s;
cout<<s;
reverse(s.begin(),s.end());
cout<<s;
return ;
}
B. Recursive Queries
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let us define two functions f and g on positive integer numbers.

You need to process Q queries. In each query, you will be given three integers lr and k. You need to print the number of integers xbetween l and r inclusive, such that g(x) = k.

Input

The first line of the input contains an integer Q (1 ≤ Q ≤ 2 × 105) representing the number of queries.

Q lines follow, each of which contains 3 integers lr and k (1 ≤ l ≤ r ≤ 106, 1 ≤ k ≤ 9).

Output

For each query, print a single line containing the answer for that query.

Examples
input

Copy
4
22 73 9
45 64 6
47 55 7
2 62 4
output
1
4
0
8
input

Copy
4
82 94 6
56 67 4
28 59 9
39 74 4
output
3
1
1
5
Note

In the first example:

  • g(33) = 9 as g(33) = g(3 × 3) = g(9) = 9
  • g(47) = g(48) = g(60) = g(61) = 6
  • There are no such integers between 47 and 55.
  • g(4) = g(14) = g(22) = g(27) = g(39) = g(40) = g(41) = g(58) = 4

他本来是递归函数,我们需要先预处理就可以了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+;
int a[N][];
int main()
{
ios::sync_with_stdio(false);
for(int i=;i<=1e6;i++)
{
int t=i;
while(t>=)
{
int s=;
while(t)
{
if(t%)s*=t%;
t/=;
}
t=s;
}
for(int j=;j<;j++)
a[i][j]=a[i-][j]+(t==j);
}
int T;
cin>>T;
while(T--)
{
int l,r,k;
cin>>l>>r>>k;
cout<<a[r][k]-a[l-][k]<<"\n";
} return ;
}
C. Permutation Cycle
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

For a permutation P[1... N] of integers from 1 to N, function f is defined as follows:

Let g(i) be the minimum positive integer j such that f(i, j) = i. We can show such j always exists.

For given N, A, B, find a permutation P of integers from 1 to N such that for 1 ≤ i ≤ Ng(i) equals either A or B.

Input

The only line contains three integers N, A, B (1 ≤ N ≤ 106, 1 ≤ A, B ≤ N).

Output

If no such permutation exists, output -1. Otherwise, output a permutation of integers from 1 to N.

Examples
input

Copy
9 2 5
output
6 5 8 3 4 1 9 2 7
input

Copy
3 2 1
output
1 2 3 
Note

In the first example, g(1) = g(6) = g(7) = g(9) = 2 and g(2) = g(3) = g(4) = g(5) = g(8) = 5

In the second example, g(1) = g(2) = g(3) = 1

递归版的轮换,一组等于a,一组等于b即可

#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,a,b,f=,fa,fb;
cin>>n>>a>>b;
if(b>a)swap(a,b);
for(int i=; i<=n&&f; i+=a)
if((n-i)%b==)
fa=i/a,fb=(n-i)/b,f=;
if(f)
cout<<-;
else
{
int i=;
for(; i<=fa*a; i+=a)
{
cout<<i+a-<<" ";
for(int j=i; j<i+a-; j++)
cout<<j<<" ";
}
for(; i<=n; i+=b)
{
cout<<i+b-<<" ";
for(int j=i; j<i+b-; j++)
cout<<j<<" ";
}
}
return ;
}
D. Tree
time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

You are given a node of the tree with index 1 and with weight 0. Let cnt be the number of nodes in the tree at any instant (initially, cnt is set to 1). Support Q queries of following two types:

  •  Add a new node (index cnt + 1) with weight W and add edge between node R and this node.
  •  Output the maximum length of sequence of nodes which
    1. starts with R.
    2. Every node in the sequence is an ancestor of its predecessor.
    3. Sum of weight of nodes in sequence does not exceed X.
    4. For some nodes i, j that are consecutive in the sequence if i is an ancestor of j then w[i] ≥ w[j] and there should not exist a node k on simple path from i to j such that w[k] ≥ w[j]

The tree is rooted at node 1 at any instant.

Note that the queries are given in a modified way.

Input

First line containing the number of queries Q (1 ≤ Q ≤ 400000).

Let last be the answer for previous query of type 2 (initially last equals 0).

Each of the next Q lines contains a query of following form:

  • 1 p q (1 ≤ p, q ≤ 1018): This is query of first type where  and . It is guaranteed that 1 ≤ R ≤ cnt and 0 ≤ W ≤ 109.
  • 2 p q (1 ≤ p, q ≤ 1018): This is query of second type where  and . It is guaranteed that 1 ≤ R ≤ cntand 0 ≤ X ≤ 1015.

 denotes bitwise XOR of a and b.

It is guaranteed that at least one query of type 2 exists.

Output

Output the answer to each query of second type in separate line.

Examples
input

Copy
6
1 1 1
2 2 0
2 2 1
1 3 0
2 2 0
2 2 2
output
0
1
1
2
input

Copy
6
1 1 0
2 2 0
2 0 3
1 0 2
2 1 3
2 1 6
output
2
2
3
2
input

Copy
7
1 1 2
1 2 3
2 3 3
1 0 0
1 5 1
2 5 0
2 4 0
output
1
1
2
input

Copy
7
1 1 3
1 2 3
2 3 4
1 2 0
1 5 3
2 5 5
2 7 22
output
1
2
3
Note

In the first example,

last = 0

- Query 1: 1 1 1, Node 2 with weight 1 is added to node 1.

- Query 2: 2 2 0, No sequence of nodes starting at 2 has weight less than or equal to 0. last = 0

- Query 3: 2 2 1, Answer is 1 as sequence will be {2}. last = 1

- Query 4: 1 2 1, Node 3 with weight 1 is added to node 2.

- Query 5: 2 3 1, Answer is 1 as sequence will be {3}. Node 2 cannot be added as sum of weights cannot be greater than 1. last = 1

- Query 6: 2 3 3, Answer is 2 as sequence will be {3, 2}. last = 2

对于一棵树,你有2种操作

ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined)的更多相关文章

  1. Codeforces 932 A.Palindromic Supersequence (ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined))

    占坑,明天写,想把D补出来一起写.2/20/2018 11:17:00 PM ----------------------------------------------------------我是分 ...

  2. ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) A

    2018-02-19 A. Palindromic Supersequence time limit per test 2 seconds memory limit per test 256 mega ...

  3. Codeforces 932 C.Permutation Cycle-数学 (ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined))

    C. Permutation Cycle   time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

  4. Codeforces 932 B.Recursive Queries-前缀和 (ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined))

    B. Recursive Queries   time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

  5. 【ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) D】Tree

    [链接] 我是链接,点我呀:) [题意] 让你在树上找一个序列. 这个序列中a[1]=R 然后a[2],a[3]..a[d]它们满足a[2]是a[1]的祖先,a[3]是a[2]的祖先... 且w[a[ ...

  6. 【ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) C】 Permutation Cycle

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] p[i] = p[p[i]]一直进行下去 在1..n的排列下肯定会回到原位置的. 即最后会形成若干个环. g[i]显然等于那个环的大 ...

  7. 【ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) B】Recursive Queries

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 写个记忆化搜索. 接近O(n)的复杂度吧 [代码] #include <bits/stdc++.h> using nam ...

  8. 【ICM Technex 2018 and Codeforces Round #463 (Div. 1 + Div. 2, combined) A】 Palindromic Supersequence

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 字符串倒着加到原串右边就好 [代码] #include <bits/stdc++.h> using namespace ...

  9. ICM Technex 2017 and Codeforces Round #400 (Div. 1 + Div. 2, combined) A map B贪心 C思路前缀

    A. A Serial Killer time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

随机推荐

  1. SequenceFile和MapFile

    HDFS和MR主要针对大数据文件来设计,在小文件处理上效率低.解决方法是选择一个容器,将这些小文件包装起来,将整个文件作为一条记录,可以获取更高效率的储存和处理,避免多次打开关闭流耗费计算资源.hdf ...

  2. 字符串(String)杂谈

    作者:臧圩人(zangweiren) 网址:http://zangweiren.javaeye.com >>>转载请注明出处!<<< 上一次我们已经一起回顾了面试题 ...

  3. .NET 读取视频文件

    该篇文章 复制别人的文章 在.NET中处理视频是一件痛苦的事情,.NET并没有提供视频处理的类.于是咱们只能找一些第三方的类库或者自己实现,在项目时间比较赶的情况下,自己实现是不可能的了,而且说不定会 ...

  4. LINQ 基础语句

    去全部集合 using (dat0216DataContext con = new dat0216DataContext()) { //LoList   是转换成  List集合 List<Us ...

  5. BZOJ 2851: 极限满月 虚树 or 树链的并

    2851: 极限满月 Time Limit: 20 Sec  Memory Limit: 512 MBSubmit: 170  Solved: 82[Submit][Status][Discuss] ...

  6. UVALive 4329 Ping pong (BIT)

    枚举中间的人,只要知道在这个人前面的技能值比他小的人数和后面技能值比他小的人数就能计算方案数了,技能值大的可有小的推出. 因此可以利用树状数组,从左到右往树上插点,每个点询问sum(a[i]-1)就是 ...

  7. 百度site网址显示完整站点信息的分析

    去年赛花红就发现百度site本博客网址,仅出现找到相关结果数约多少个,数字为估算值,网站管理员如需了解更准确的索引量,请使用百度站长平台等字样.但赛花红又发现有的网站却显示着完整的站点信息,当时以为是 ...

  8. TDB文件介绍

    samba在运行时,Samba 存储许多信息,从本地密码到希望从中收到信息的一系列客户端.这类数据其中一些是暂时的,在 Samba 重启时可能会被丢弃,但是另一些却是永久的,不会被丢弃.这类数据可能是 ...

  9. $|^|\z|\Z|/a|/l

    #!/usr/bin/perl use strict; use warnings; foreach(<>) { if (/(\w*)/a){print "$1\n";} ...

  10. PAT (Basic Level) Practise (中文)- 1018. 锤子剪刀布 (20)

    http://www.patest.cn/contests/pat-b-practise/1018 大家应该都会玩“锤子剪刀布”的游戏:两人同时给出手势,胜负规则如图所示: 现给出两人的交锋记录,请统 ...