[Blue Bridge Cup] [previous questions] turn the coin Easy only once * greedy issues

The basic idea:

Greedy grasp the idea is not very good, in the case of missing an enumeration:

010

111

At this time, twice requiring inversion;

 

If you use a greedy strategy, it is the moment to find the optimal solution, namely: direct encounter different direct translation, followed by a subsequent judgment;

 

key point:

no;

 

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<vector> 
#include<string>
#include<math.h>
#include<algorithm>
#include<cstring>
#include<map>
#include<queue>
#include<set>
#include<stack>
using namespace std;


int main() {
    string s1, s2;
    CIN >> >> S1 S2;
     int CNT = 0 ;
     for ( int I = 0 ; I <s1.size (); I ++ ) {
         IF (! S1 [I] = S2 [I]) {
             // when encountering when not equal; 
            IF (S1 [I + . 1 ] == ' * ' )
                S1 [i + 1 ] = ' O ' ;
            presence 
                S1 [i + 1 ] = ' * ' ;
            cnt++;
        }
    }
    cout << cnt;
    return 0;
}