Problem:
Score : 300 points

Problem Statement

Given is a string S. Let T be the concatenation of K copies of S.
We can repeatedly perform the following operation: choose a character in T and replace it with a different character.
Find the minimum number of operations required to satisfy the following condition: any two adjacent characters in T are different.

Constraints

1 \leq |S| \leq 100

S consists of lowercase English letters.

1 \leq K \leq 10^9

K is an integer.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the minimum number of operations required.

Sample Input 1

issii
2

Sample Output 1

4

T is issiiissii. For example, we can rewrite it into ispiqisyhi, and now any two adjacent characters are different.

Sample Input 2

qq
81

Sample Output 2

81

Sample Input 3

cooooooooonteeeeeeeeeest
999993333

Sample Output 3

8999939997

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

int main()
{
    long long n,m,p,q,r,i,j,k,x,y,z;
    string s;
    cin>>s>>k;
    p=0;
    if(s[0]==s[s.size()-1])
    {
        p=1;
    }
    if(s[s.size()-1]==s[s.size()-2])
    {
        y=2;
    }

    if(s.size()<=3)
    {
        if(p==1 && y==2 && s.size()==3)
        {
            cout<<k*2<<endl;
            return 0;
        }
        else if(s.size()==3 && p==1)
        {
            cout<<k-1<<endl;
            return 0;
        }
        else if(s.size()==3 && y==2)
        {
            cout<<k<<endl;
            return 0;
        }
        else if(p==1 && y==2 && s.size()==2)
        {
            cout<<k<<endl;
            return 0;
        }
        else if(s.size()==1)
        {
            cout<<k/2<<endl;
            return 0;
        }
        else
        {
            cout<<0<<endl;
            return 0;
        }
    }

    if(y==2)
    {
        x=0;
        for(i=0; i<s.size()-2; i++)
        {
            if(s[i]==s[i+1])
            {
                x=x+2;
                i++;
            }
        }
        x=x/2;
        q=k+(k*x);
        cout<<q<<endl;
    }
    else if(p==1)
    {
        x=0;
        for(i=0; i<s.size()-2; i++)
        {
            if(s[i]==s[i+1])
            {
                x=x+2;
                i++;
            }
        }
        x=x/2;
        q=(k-1)+(k*x);
        cout<<q<<endl;
    }
    else
    {
        x=0;
        for(i=0; i<s.size()-1; i++)
        {
            if(s[i]==s[i+1])
            {
                x=x+2;
                i++;
            }
        }
        //cout<<x<<endl;
        x=x/2;
        //cout<<x<<endl;
        cout<<k*x<<endl;
    }
}