题解 | abb

水题，考虑直接倒序更新，一边更新一边维护 每种字符的后缀数量和 suf[c]，然后动态更新sum = $\sum (suf[c] - 1) * suf[c] / 2$，最后更新答案 ans += sum - sum (suf[s[i]] - 1) * suf[s[i]] / 2，即可。

#include <bits/stdc++.h>
using i64 = long long;

int main() {
    std::cin.tie(nullptr)->sync_with_stdio(false);

    int n;
    std::string s;
    std::cin >> n >> s;

    std::vector<int> suf(26);
    i64 sum = 0;

    i64 ans = 0;
    for (int i = n - 1; i >= 0; i--) {
        ans += sum - (suf[s[i] - 'a'] - 1) * suf[s[i] - 'a'] / 2;
        
        sum -= (suf[s[i] - 'a'] - 1) * suf[s[i] - 'a'] / 2;
        suf[s[i] - 'a']++;
        sum += (suf[s[i] - 'a'] - 1) * suf[s[i] - 'a'] / 2;
    }

    std::cout << ans;

    return 0;
}

https://www.nowcoder.com/discuss/726480854079250432