A. Kyoya and Photobooks

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/554/problem/A

Description

Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?

Please help Haruhi solve this problem.

Input

The first line of input will be a single string s (1 ≤ |s| ≤ 20). String s consists only of lowercase English letters.

Output

Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.

Sample Input

a

Sample Output

51

HINT

题意

给你一个字符串,然后让你随便在一个位置加上一个字母,问你能够产生多少个不同的字符串

题解:

暴力加字符串,然后用map判重就好了

代码

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/* inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH[Num--]+48);
puts("");
}
*/
inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** map<string,int> H; int main()
{
string s;
cin>>s;
int ans=;
for(int i=;i<=s.size();i++)
{
for(int j=;j<;j++)
{
string s1;
if(i==)
s1=char(j+'a')+s;
else if(i==s.size())
s1=s+char(j+'a');
else
s1=s.substr(,i)+char(j+'a')+s.substr(i,s.size());
if(!H[s1])
{
ans++;
H[s1]=;
}
}
}
cout<<ans<<endl;
}

Codeforces Round #309 (Div. 2) A. Kyoya and Photobooks 字符串水题的更多相关文章

  1. Codeforces Round #309 (Div. 2) B. Ohana Cleans Up 字符串水题

    B. Ohana Cleans Up Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/554/pr ...

  2. 找规律 Codeforces Round #309 (Div. 2) A. Kyoya and Photobooks

    题目传送门 /* 找规律,水 */ #include <cstdio> #include <iostream> #include <algorithm> #incl ...

  3. Codeforces Round #309 (Div. 2) A. Kyoya and Photobooks【*组合数学】

    A. Kyoya and Photobooks time limit per test 2 seconds memory limit per test 256 megabytes input stan ...

  4. Codeforces Round #368 (Div. 2) A. Brain's Photos (水题)

    Brain's Photos 题目链接: http://codeforces.com/contest/707/problem/A Description Small, but very brave, ...

  5. Codeforces Round #373 (Div. 2) C. Efim and Strange Grade 水题

    C. Efim and Strange Grade 题目连接: http://codeforces.com/contest/719/problem/C Description Efim just re ...

  6. Codeforces Round #185 (Div. 2) A. Whose sentence is it? 水题

    A. Whose sentence is it? Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/ ...

  7. Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 水题

    A. Vitya in the Countryside 题目连接: http://codeforces.com/contest/719/problem/A Description Every summ ...

  8. Codeforces Round #371 (Div. 2) A. Meeting of Old Friends 水题

    A. Meeting of Old Friends 题目连接: http://codeforces.com/contest/714/problem/A Description Today an out ...

  9. Codeforces Round #355 (Div. 2) B. Vanya and Food Processor 水题

    B. Vanya and Food Processor 题目连接: http://www.codeforces.com/contest/677/problem/B Description Vanya ...

随机推荐

  1. Delphi 调用串口例子

    procedure TfrmClientMain.SayAddr;var sbuf:array[1..7] of byte;begin sbuf[1]:=byte($35); sbuf[2]:=byt ...

  2. Python编程中的反模式

    Python是时下最热门的编程语言之一了.简洁而富有表达力的语法,两三行代码往往就能解决十来行C代码才能解决的问题:丰富的标准库和第三方库,大大节约了开发时间,使它成为那些对性能没有严苛要求的开发任务 ...

  3. 哈希(Hash)与加密(Encrypt)的基本原理、区别及工程应用

    0.摘要 今天看到吉日嘎拉的一篇关于管理软件中信息加密和安全的文章,感觉非常有实际意义.文中作者从实践经验出发,讨论了信息管理软件中如何通过哈希和加密进行数据保护.但是从文章评论中也可以看出很多朋友对 ...

  4. mysql执行update报错1175解决方法

    mysql执行update报错 update library set status=true where 1=1 Error Code: 1175. You are using safe update ...

  5. Android调用系统自带的文件管理器进行文件选择并读取

    先调用: intent = new Intent(Intent.ACTION_GET_CONTENT); intent.setType("*/*"); //设置类型,我这里是任意类 ...

  6. 深入浅出谈存储之NAS是什么

    深入浅出谈存储之NAS是什么 2012年02月17日16:42 来源:新浪博客 作者:林沛满 编辑:曾智强 查看全文 赞(0)评论(0) 分享 标签: NAS , 企业NAS , 存储系统 [IT16 ...

  7. STM32 常用GPIO操作函数记录

    STM32读具体GPIOx的某一位是1还是0 /** * @brief Reads the specified input port pin. * @param GPIOx: where x can ...

  8. HDU 5927 Auxiliary Set (dfs)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5927 题意: 给你一棵树,其中有一些'不重要'的点,要是这些'不重要'的点的子树中有两个重要的点的LC ...

  9. HDU 2196Computer(树形DP)

    给你一颗边带权值的树,求树上的每一点距离其最远的一个点的距离 比较典型的题了,主要方法是进行两次DFS,第一次DFS求出每一个点距离它的子树的最远距离和次远距离,然后第二次DFS从父节点传过来另一侧的 ...

  10. C# JackLib系列之GdiHelper圆角矩形的快速生成

    using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.D ...