Title description
Xiao Ming is playing a "coin flip" game.
There are several coins in a row on the table. We use * to represent the front and o to represent the back (lowercase letters, not zeros).
For example, the possible situations are:
If you flip the two coins on the left at the same time, it becomes:
Now Xiaoming’s question is: If the initial state and the target state to be reached are known, and only two adjacent coins can be flipped at the same time at a time, how many times is the least flipped for a particular situation?
We agree: Turning two adjacent coins is called a one-step operation.
Input format
Two lines of equal length character strings respectively represent the initial state and the target state to be reached. (The length of each line <1000)
Output format
An integer representing the minimum number of operation steps.
Sample input 
Sample output
1
Thinking analysis
Define two strings by comparing and replacing them so that they become two identical strings.
Add a counter to get the result
- The source code is as follows
#include<stdio.h>
#include<string.h>
int main()
{
int cnt = 0;
char s1[1000];
char s2[1000];
scanf("%s%s", s1, s2);
for(int i = 0; i < strlen(s1); i++){
if(s1[i] != s2[i]){
if(s2[i+1] == '*')
s2[i+1] = 'o';
else if(s2[i+1] == 'o')
s2[i+1] = '*';
cnt ++;
}
}
printf("%d", cnt);
return 0;
}
problem solved.