Codeforces Round #309 (Div. 2) A. Kyoya and Photobooks【*组合数学】
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.
The first line of input will be a single string s (1 ≤ |s| ≤ 20). String s consists only of lowercase English letters.
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
a
51
hi
76
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
【题意】： 给一个长度不大于20的字符串。任意一个位置插入一个a-z 。求一共有多少新字符串的可能。
【分析】：长度为n的字符串 有n+1个空位。第一个位置有26种。剩下的位置， 只需要和前一个位置避免重复就好，所以都是25种。所以答案就是26+25*n。
【代码】：
#include <bits/stdc++.h>
using namespace std;
char a[25];
int main()
{
int n;
scanf("%s",a);
n=strlen(a);
printf("%d
",n*25+26);
}