1 solutions

  • 0
    @ 2025-11-5 20:06:40

    C++ :

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    //#include<>
    using namespace std;
    int main()
    {
    	int l,r,tot=0;
    	cin>>l>>r;
    	for(int i=l;i<=r;i++)
    	{
    		int x=i;
    		while(x>0)
    		{
    			if(x%10==2) tot++;
    			x=x/10;
    		}
    	}
    	cout<<tot<<endl;
    	return 0;
    }
    

    Pascal :

    var
      l,r:1..10000;
      i,j,h,c:longint;
      s:string;
    begin
      {assign(input,'two.in');
      assign(output,'two.out');
      reset(input);
      rewrite(output);}
      readln(l,r);
      c:=0;
      for i:=l to r do
        begin
          str(i,s);
          h:=length(s);
          for j:=1 to h do
            if s[j]='2'
              then c:=c+1;
        end;
      writeln(c);
     { close(input);
      close(output);}
    end.
    

    Java :

    import java.util.Scanner;
    
    
    public class Main {
    
    	public static void main(String[] args) {
    		Scanner sc = new Scanner(System.in);
    		while(sc.hasNextLine()){
    			String s = sc.nextLine();
    			String[] ss = s.split(" ");
    			if(ss.length!=2){
    				return;
    			}
    			int a = Integer.parseInt(ss[0]);
    			int b = Integer.parseInt(ss[1]);
    			int count1 = getCount(a),count2 = getCount(b);
    			
    			while(a!=0){
    				int t = a%10;
    				if(t==2){
    					count1 --;
    				}
    				a = a/10;
    			}
    			System.out.println(count2-count1);
    			
    		}
    		
    		
    		
    	}
    
    	public static int getCount(int n){
    		int count = 0;
    		int k = 1;
    		while(k<=n){
    			int after = n%k;
    			int before  = n/10/k;
    			int cur = n/k%10;
    			
    			if(cur>2){
    				count = count+ (before+1)*k;
    			}else if(cur<2){
    				count = count + before*k;
    			}else if (cur==2){
    				count = count + before*k+after+1;
    			}
    			k*=10;
    
    			
    		}
    		
    		
    		
    		return count;
    	}
    }
    
    
    • 1

    Information

    ID
    20017
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
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    Accepted
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