Codeforce c.strange test and acwing

C.Strange Test— greedy + An operation

Answer key ：

First we need to know ,a|b It must be greater than or equal to b Of , After using operation three, only operation two , Therefore, operation three can only be used once , So this problem turns into a++,b++, When two numbers are added to a value, they make a|b=b, Finding the value with the least number of operation steps is the answer
It's not difficult to think about this problem , It's really hard to knock code if you're not familiar with bit operation

An operation

#include <iostream>
using namespace std;
signed main()
{
    
    int T;
    int a,b;
	cin>>T;
    while(T--)
    {
    
    	scanf("%d %d",&a,&b);
    	int res=1e6+2;
    	res=b-a;
    	for(int i=a;i<=b;i++)
    	{
    
    		int x=b;
    		for(int j=20;j>=0;j--)
    		{
    
    		// Find the first a|b==b Of b value 
    			if(i>>j&1)
    			{
    
    				if(~x>>j&1)
    				{
    
    					x|=1<<j;
    					x&=~((1<<j)-1);
					}
				}
			}
			res=min(res,i-a+1+x-b);
		}
		cout<<res<<endl;
	}
}

acwing Weekly game Two kinds of operations

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int n,m;
int res=0;
signed main()
{
    
	cin>>n>>m;
	while(m>n)
	{
    
	    if(m&1)res++;
	    m=(m+1)/2;
	    res++;
	}
	res+=n-m;
	cout<<res<<endl;
}