//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
7
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 ",sum); 44 45 } 46 47 return 0; 48 49 }