A. Kyoya and Photobooks

Time Limit: 20 Sec

Memory Limit: 256 MB

題目連接

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

?

題意

給你一個字符串，然后讓你隨便在一個位置加上一個字母，問你能夠產(chǎn)生多少個不同的字符串

題解：

暴力加字符串，然后用map判重就好了

代碼

?

//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[20]; //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=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f; } 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(""); } //************************************************************************************** map<string,int> H;int main() {string s;cin>>s;int ans=0;for(int i=0;i<=s.size();i++){for(int j=0;j<26;j++){string s1;if(i==0)s1=char(j+'a')+s;else if(i==s.size())s1=s+char(j+'a');elses1=s.substr(0,i)+char(j+'a')+s.substr(i,s.size());if(!H[s1]){ans++;H[s1]=1;}}}cout<<ans<<endl; }

?

轉(zhuǎn)載于:https://www.cnblogs.com/qscqesze/p/4599068.html

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