Codeforces Round #258 (Div. 2) D. Count Good Substrings —— 组合数学

题目链接：http://codeforces.com/problemset/problem/451/D

D. Count Good Substrings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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 length n (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.

Examples

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 substring s[l, r] (1 ≤ l ≤ r ≤ n) of
string s = s1s2... sn is
string slsl + 1... sr.

A string s = s1s2... sn is
a palindrome if it is equal to string snsn - 1... s1.

题解：

1.分别记录‘a’在奇数位置、偶数位置出现的次数， ‘b’亦如此。

2.长度为偶数的情况：相同的字母一个出现在奇数位置， 一个出现在偶数位置，假设出现次数分别为n、m， 则n*m。

3.长度为偶数的情况：相同的字母都出现在偶数位置或者奇数位置。

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 1e5+10;

char s[maxn];
int a[2][2];

LL f(int x) { return 1LL*x*(x-1)/2; }

int main()
{
    scanf("%s",s+1);
    int len = strlen(s+1);
    for(int i = 1; i<=len; i++)
        a[s[i]!='a'][i&1]++;

    LL even = 1LL*a[0][0]*a[0][1] + 1LL*a[1][0]*a[1][1];
    LL odd = f(a[0][0]) + f(a[0][1]) + f(a[1][0]) + f(a[1][1]) + len;
    cout<<even<<' '<<odd<<endl;
}