Educational Codeforces Round 74 (Rated for Div. 2) D. AB-string

链接:

https://codeforces.com/contest/1238/problem/D

题意:

The string t1t2…tk is good if each letter of this string belongs to at least one palindrome of length greater than 1.

A palindrome is a string that reads the same backward as forward. For example, the strings A, BAB, ABBA, BAABBBAAB are palindromes, but the strings AB, ABBBAA, BBBA are not.

Here are some examples of good strings:

t = AABBB (letters t1, t2 belong to palindrome t1…t2 and letters t3, t4, t5 belong to palindrome t3…t5);

t = ABAA (letters t1, t2, t3 belong to palindrome t1…t3 and letter t4 belongs to palindrome t3…t4);

t = AAAAA (all letters belong to palindrome t1…t5);

You are given a string s of length n, consisting of only letters A and B.

You have to calculate the number of good substrings of string s.

思路:

考虑ABB这种无法满足, 可以推出一只有一个a或b且处于子串的左右两边的时候,不满足, 记录不满足的个数,减一下.

代码:

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 3e5+10;

char s[MAXN];
int n;

int main()
{
        cin >> n;
        cin >> (s+1);
        LL cnt = 0;
        for (int i = 1, j;i < n;i = j)
        {
                j = i;
                while (j+1 <= n && s[j+1] == s[i])
                        j++;
                i = j;
                while (j+1 <= n && s[j+1] != s[i])
                        j++;
                cnt += j-i;
        }
        for (int i = n, j;i > 1;i = j)
        {
                j = i;
                while (j-1 >= 1 && s[j-1] == s[i])
                        j--;
                i = j;
                while (j-1 >= 1 && s[j-1] != s[i])
                        j--;
                if (i-j-1 > 0)
                        cnt += i-j-1;
        }
        LL ans = (1LL*n*(n-1))/2-cnt;
        cout << ans << endl;

        return 0;
}