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(){
  string S;
  int long long K;
  cin>>S>>K;
  int len = S.size();
  int x;
  int long long ans=0;
  if(len==1){
    cout<<K/2<<endl;
    exit(0);
  }
  if(len==2){
    if(S.at(0)==S.at(1)){
      cout<<K<<endl;
      exit(0);
    }
    else{
      cout<<0<<endl;
      exit(0);
    }
  }
  if(S.at(0)==S.at(len-1))
    x=1;
  else
    x=0;
  int num=0;
  if(x==0){
    for(int i=1;i<len;i++){
      if(S.at(i)==S.at(i-1))
        num++;
      else{
        ans+=((num+1)/2)*K;
        num=0;
      }
    }
    cout<<ans<<endl;
  }
  int rnum=0;
  int s=0;
  int e;
  if(x==1){
    for(int i=1;i<len-1;i++){
      if(S.at(i)==S.at(0))
        rnum++;
      else{
        s=i;
        break;
      }
    }
    if(rnum==len-2){
      ans=(len*K)/2;
      cout<<ans<<endl;
    }
    else{
      for(int i=len-2;i>0;i--){
        if(S.at(i)==S.at(len-1))
          rnum++;
        else{
          e=i;
          break;
        }
      }
      for(int i=s;i<e+2;i++){
        if(S.at(i)==S.at(i-1))
          num++;
        else{
          ans+=((num+1)/2)*K;
          num=0;
        }
      }
      rnum+=2;
      ans+=(rnum/2)*K;
      cout<<ans<<endl;
    }
  }
}