（寒假CF）Choosing Symbol Pairs

//我觉得这题好像跟我理解的不太一样啊T ^ T，有点郁闷

Description

There is a given string S consisting of N symbols. Your task is to find the number of ordered pairs of integers i and j such that

1. 1 ≤ i, j ≤ N

2. S[i] = S[j], that is the i-th symbol of string S is equal to the j-th.

Input

The single input line contains S, consisting of lowercase Latin letters and digits. It is guaranteed that string S in not empty and its length does not exceed 105.

Output

Print a single number which represents the number of pairs i and j with the needed property. Pairs (x, y) and (y, x) should be considered different, i.e. the ordered pairs count.

Sample Input

Input

great10

Output

Input

aaaaaaaaaa

Output

100

 1 #include<stdio.h>
 2
 3 #include<string.h>
 4
 5
 6
 7 int t[100000];
 8
 9
10
11 int main()
12
13 {
14
15 char s[100000];
16
17 while(gets(s))
18
19 {
20
21 double sum=0;
22
23 int l=strlen(s);
24
25 memset(t,0,sizeof(t));
26
27 for(int i=0;i<l;i++)
28
29 {
30
31 t[s[i]]++;
32
33 }
34
35 for(int i='0';i<='z';i++)
36
37 {
38
39 sum+=(double)t[i]*t[i];
40
41 }
42
43 printf("%.0lf\n",sum);
44
45 }
46
47 return 0;
48
49 }

View Code

转载于:https://www.cnblogs.com/awsent/p/4266900.html

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