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51nod 1413 power binary

2022-07-18 03:19:00 【51CTO】




A decimal integer is called a power binary , When expressed in decimal, it is only represented by 0 or 1 form . for example 0,1,101,110011 It's all power binary and 2,12,900 No .

When given a n When , Calculate the minimum number of power binaries you need to add to get n.




Input



A single set of test data . The first line gives an integer n (1<=n<=1,000,000)



Output



Output the answer in one line .



Input Example



9



Output Example



9



At the beginning of doing this, I took out my ancestral DFS, But it didn't come out .

Then observe the data , Find that what you give n Everyone in the group took it out , Then find someone to follow n Power binary numbers of the same length ,n If each bit of is greater than 1, The bit of this binary number is equal to 1, Or equal to 0, Then this number can definitely be added to n Of , then n Subtract this number , Repeat these steps , You can figure it out .

The code is as follows ：

      
      

       
       #include<cstdio>
       
       

       
       #include<cstring>
       
       

       
       #include<iostream>
       
       

       
       #include<algorithm>
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       #define N 300
       
       

       
       #define inf 0xfffffff
       
       

       
       int 
       
       n,
       
       Min;
       
       

       
       int 
       
       main(){
       
       
    
       
       int 
       
       i,
       
       j,
       
       m,
       
       r,
       
       x;
       
       
    
       
       scanf(
       
       "%d",
       
       &
       
       n);
       
       
    
       
       int 
       
       cnt
       
       =
       
       0,
       
       q;
       
       
    
       
       j
       
       =
       
       1;
       
       
    
       
       while(
       
       n){
       
       
        
       
       int 
       
       len
       
       =
       
       0;
       
       
       
       
       r
       
       =
       
       n;
       
       
      
       
       while(
       
       r){
       
       
        
       
       r
       
       /=
       
       10;
       
       
        
       
       len
       
       ++;
       
       
      }
       
       
      
       
       r
       
       =
       
       n,
       
       j
       
       =
       
       1,
       
       q
       
       =
       
       0;
       
       
      
       
       for(
       
       i
       
       =
       
       1;
       
       i
       
       <=
       
       len;
       
       i
       
       ++){
       
       
          
       
       int 
       
       x
       
       =(
       
       r
       
       /
       
       j)
       
       %
       
       10;
       
       
          
       
       if(
       
       x
       
       >=
       
       1) 
       
       x
       
       =
       
       1;
       
       
          
       
       else 
       
       x
       
       =
       
       0;
       
       
          
       
       q
       
       +=(
       
       x
       
       *
       
       j);
       
       
          
       
       j
       
       *=
       
       10;
       
       
      }
       
       
      
       
       n
       
       -=
       
       q;
       
       
      
       
       cnt
       
       ++;
       
       
    }
       
       
    
       
       printf(
       
       "%d\n",
       
       cnt);
       
       
    
       
       return 
       
       0;
       
       
}
      
      

      
      
    
       
       
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