看到最小操作步数就想到了贪心 然后开始找怎么贪心 原理很好想 就看代码吧

#include<bits/stdc++.h>
using namespace std;
char a[10000];
char b[10000];
int main(){
	cin>>a>>b;
	int num=0;
	int sum=0;
	int num1,num2;
	for(int i=0;i<strlen(a);i++){
		if(a[i]!=b[i]){
			num++;
			if(num==1){
				num1=i;
			}
			else if(num==2){
				num2=i;
				sum=num2-num1+sum;
				num=0;
			}
		}
	}
	cout<<sum;


	return 0;

}

贪心没什么好说的

#include<bits/stdc++.h>
using namespace std;
int ans; 
int main()
{
	std::ios::sync_with_stdio(false);
	string start; 
	string finish;
	cin>>start>>finish;
	for(int i=0;i<start.length();i++)
	{
		if(start[i]!=finish[i])
		{
			ans++;
			if(start[i]=='*')//i不一样就变化 
				start[i]='o';
			else
				start[i]='*';
			
			if(start[i+1]=='*')//i+1同时变化 
				start[i+1]='o';
			else
				start[i+1]='*';
		}
	}
	cout<<ans;
	return 0;
}

#include<iostream>
#include<cstring>
using namespace std;
const int N=1007;
char a[N],b[N];
int ans;
int main(){
	cin>>a>>b;
	int n=strlen(a); 
	for(int i=0;i<n-1;i++){
		if(a[i]==b[i])continue;
		else if(a[i]=='*'){
			ans++;
			if(a[i+1]=='*')a[i+1]='o';
			else a[i+1]='*';
		}
		else if(a[i]=='o'){
			ans++;
			if(a[i+1]=='*')a[i+1]='o';
			else a[i+1]='*';
		}
	}
	cout<<ans;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define accelerate ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define endl "\n";
#define mod 1000000007
int main(){
	accelerate;
	string s;
	string ss;
	int ans=0;
	cin>>s>>ss;
	int i=0,j=0;
	while(i<s.size()-1){
		if(s[i]!=ss[j]){
			ans++;
			if(s[i+1]=='*') s[i+1]='o';
			else s[i+1]='*';
		}
		i++;j++;
	}
	cout<<ans;
	return 0;
}

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<map>


using namespace std;

char arr1[1005];
char arr2[1005];

void op(int n)//函数操作两个字符串
{
	if (arr2[n] == '*') arr2[n] = 'o';
	else arr2[n] = '*';
	if (arr2[n+1] == '*') arr2[n+1] = 'o';
	else arr2[n+1] = '*';
}

int main(void)
{
	cin.getline(arr1, 1000).getline(arr2, 1000);
//数据输入
	int len = strlen(arr1);
	int ans = 0;
	for (int i = 0; i < len; i++)
	{
		if (arr1[i] != arr2[i])
		{
			ans++;
			op(i);
		}

	}
	cout << ans;
//思想就是模拟翻硬币的过程
	return 0;
}