Choosing Symbol Pairs

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 ≤ i, j ≤ N
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
题目很水，就是让你配对，题意也很好理解，注意int不行，会越界,我错了两次就是没有进行好强制转换，看一下我的代码吧

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <algorithm>
using namespace std;
int main()
{char x[100005];int a[50];while(~scanf("%s", x)){memset(a, 0, sizeof(a));int len = strlen(x);for(int i = 0; i < len; i++){if(islower(x[i])) a[x[i] - 'a' + 10]++;else a[x[i] - '0']++;}long long sum = 0;for(int i = 0; i < 36; i++){sum += (a[i] * 1LL *a[i]);}printf("%lld\n", sum);}return 0;
}

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