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.

Examples
Input
a
Output
51
Input
hi
Output
76
Note
In the first case, we can make ‘a(chǎn)b’,‘a(chǎn)c’,…,‘a(chǎn)z’,‘ba’,‘ca’,…,‘za’, and ‘a(chǎn)a’, producing a total of 51 distinct photo booklets.
原串長(zhǎng)度為L(zhǎng)，則共有L + 1個(gè)位置可以插入，每個(gè)位置可以有26種插法，共26*(L + 1)種。但注意相鄰兩個(gè)空位插入同為位于這兩個(gè)相鄰空位中間的字母，構(gòu)造出的新串是一樣的，所以答案要減去L，于是最后答案為25*L + 26

C語(yǔ)言版本一

#include <stdio.h> #include <stdlib.h> #include <string.h> /* run this program using the console pauser or add your own getch, system("pause") or input loop */int main(int argc, char *argv[]) {char s[25];scanf("%s",&s);printf("%d\n",26*(strlen(s)+1)-strlen(s)) ;return 0; }

C++版本一

#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <string> #include <algorithm> #include <stack> #include <queue> #include <vector> #include <map> #include <set> using namespace std;int main() {char s[30];while(scanf("%s", s) != EOF)printf("%d\n", strlen(s)*25 + 26);return 0; }

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