写法2：乘法原理$O(n)$, 考虑这个字母被多少个子串包含，即贡献

详细题解

#include <iostream>
#include <cstring>

using namespace std;

typedef long long LL;

int last[127];

int main()
{
    string s;
    cin >> s;

    int n = s.size();
    s = '0' + s;

    LL ans = 0;
    for(int i = 1;i <= n;i ++ ){
        ans += (LL)(i - last[s[i]]) * (n - i  + 1);
        last[s[i]] = i;
    }

    cout << ans << endl;

    return 0;
}

写法1：直接暴力，50分代码

#include <iostream>
#include <cstring>

using namespace std;

const int N = 100010;

bool cnt[26];

int main()
{
    string s;
    cin >> s;

    long long res = 0;
    for(int i = 0;i < s.size();i ++ )
        for(int j = 0;j <= i;j ++ )
        {
            int sc = 0;
            memset(cnt,0,sizeof cnt);
            for(int k = j;k <= i;k ++ ) cnt[s[k] - 'a'] = 1 ;

            for(int k = 0;k < 26;k ++ ){
                if(cnt[k]) sc ++;
            }
            res += sc;
        }
    cout << res <<endl;

    return 0;
}