题解 | abb

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
typedef long long ll;
ll pro[26][N],nex[26][N];

int main() {
    int n;cin>>n;
    string str;cin>>str;
    str = ' ' + str;
    for(int i = 1;i<=n;i++){
        for(int j = 0;j<26;j++) pro[j][i] = pro[j][i-1];
        pro[str[i]-'a'][i]++;
    }

    for(int i = n;i>=1;i--){
        for(int j = 0;j<26;j++) nex[j][i] = nex[j][i+1];
        nex[str[i]-'a'][i]++;
    }

    long long ans = 0;
    for(int i = 2;i<n;i++){
        int c = str[i] -'a';
        for(int j  = 0;j<26;j++){
            if(j==c) continue;
            ans += pro[j][i] * nex[c][i+1];
        }
    }

    cout<<ans;
   

    return 0;
}
// 64 位输出请用 printf("%lld")