Codeforces round #258 (div2) d Problem Solving report

D. Count good substringstime limit per Test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

We call a string good, if after merging all the consecutive equal characters, the resulting string is palindrome. for example, "aabba" is good, because after the merging step it will become "ABA ".

Given a string, you have to find two values:

1. The number of good substrings of even length;
2. The number of good substrings of odd length.

Input

The first line of the input contains a single string of LengthN(1? ≤?N? ≤? 105). Each character of the string will be either 'A' or 'B '.

Output

Print two space-separated integers: the number of good substrings of even length and the number of good substrings of odd length.

Sample test (s) Input

bb

Output

1 2

Input

baab

Output

2 4

Input

babb

Output

2 5

Input

babaa

Output

2 7

Note

In example 1, there are three good substrings ("B", "B", and "BB"). One of them has even length and two of them have odd length.

In Example 2, there are six good substrings (I. e. "B", "A", "A", "B", "AA", "Baab "). two of them have even length and four of them have odd length.

In Example 3, there are seven good substrings (I. e. "B", "A", "B", "B", "BB", "Bab", "babb "). two of them have even length and five of them have odd length.

Definitions

A substringS[L,?R] (1? ≤?L? ≤?R? ≤?N) Of stringS? =?S1S2...SNIs stringSLSL? +? 1...SR.

A stringS? =?S1S2...SNIs a palindrome if it is equal to stringSNSN? -? 1...S1.

Question:

A string containing only a and B. Now we define a string. If the merged string is a return string, it is a good substrings. Then we can find the number of such strings and output the numbers of even and odd numbers.

Solution:

First, we need to note two known conditions:

1. Strings can be merged. For example, after merging abbaabbb, It is Abab.

2. Only two characters, A and B

We can find that the merged strings must be of the ABA or Abab type. If the merged strings are in the background, the first character must be the same as the last character, and vice versa.

We can further draw a conclusion: the strings between two different positions with the same characters must be good substrings. The total number is easily obtained at the time complexity of O (N), but we now need to split the numbers with odd and even numbers in length, then we can split the method for finding the total number.

A good substrings with an even number of characters with an odd position,

A good substrings with an odd number of characters,

A character with an even position, and a character with an even position constitute a good substrings with an odd length,

A good substrings with an even number of characters with an odd position,

Code:

#include <string>#include <iostream>#define LL long longusing namespace std;string st;LL ansOdd, ansEven;LL Odd[2], Even[2];void init() {cin >> st;}void solve() {for (int i = 0; i < st.size(); i++) {ansOdd++;int j = st[i]-'a';if ((i+1)&1) {ansOdd += Odd[j];ansEven += Even[j];Odd[j]++;}else {ansEven += Odd[j];ansOdd += Even[j];Even[j]++;}}cout << ansEven << ' ' << ansOdd << endl;}int main() {init();solve();}

Codeforces round #258 (div2) d Problem Solving report