题解 | abb,

#include <iostream>
#include<string>
using namespace std;
int pre[26];
int all[26];
int main()
{   
int n;cin>>n;string str;cin>>str;
for(char &c:str)all[c-'a']++;
long long ans=0;
for(int i=0;i<n;i++)
{
    char c=str[i];
    pre[c-'a']++;
ans+=(all[c-'a']-pre[c-'a'])*(i+1-pre[c-'a']);
}
cout<<ans;
    return 0;
}