思路：模拟 O(n)(n为字符串长度)

#include<iostream>
#include<algorithm>
#include<string>

using namespace std;

int main()
{
    string s1,s2;
    cin>>s1>>s2;
    int cnt=0;
    for(int i=0;i<s1.length();i++){
        if(s1[i]!=s2[i]){
            s1[i]=s1[i]=='*'?'o':'*';
            s1[i+1]=s1[i+1]=='*'?'o':'*';
            cnt++;
        }
    }
    cout<<cnt<<endl;
    return 0;
}