9 solutions

  • 4
    @ 2024-1-27 11:44:01
    #include<iostream>
    #include<iomanip>
    #include<cmath>
    using namespace std;
    long l,r,s=0;
    int main(){
    	cin>>l>>r;
    	for(int i=l;i<=r;i++){
    		long x=i;
    		while(x){
    			if(x%10==2) s++;
    			x/=10;
    		}
    	}
    	cout<<s;
    	return 0;
    }
    
    • 2
      @ 2023-9-22 19:53:48
      #include <bits/stdc++.h>
      using namespace std;
      int c2(int n){
      	int sum = 0;
      	while(n > 0){
      		if (n % 10 == 2){
      			sum++;
      		}
      		n /= 10;
      	}
      	return sum;
      }
      int main(int argc, char **argv){
      	int l,r,sum = 0;
      	cin >> l >> r;
      	for (int i = l;i <= r;i++){
      		sum += c2(i);
      	}
      	cout << sum;
      	return 0;
      }
      
      • 2
        @ 2023-9-18 22:08:48
        //简单题,不纠结
        #include <bits/stdc++.h>
        using namespace std;
        int l,r;
        int count(int x){
        	int cnt=0;
        	while(x){
        		int k=x%10;
        		if(k==2) cnt++;
        		x/=10;
        	}
        	return cnt;
        }
        int main(){
        	cin>>l>>r;
        	int ans=0;
        	for(int i=l;i<=r;i++) ans+=count(i);
        	cout<<ans<<endl;
        	return 0;
        }
        
        • 1
          @ 2023-9-16 16:50:51

          `

          #include <iostream> 
          using namespace std; 
          int main() { 
          int l,r; 
          int s=0; 
          cin>>l>>r; 
          for(int i=l;i<=r;i++) { 
          int q=i; 
          while(q) { 
          if(q%10==2) s++; 
          q/=10; 
          } 
          } 
          cout<<s; 
          }
          
          • 0
            @ 2024-10-20 15:01:55
            #include<bits/stdc++.h>
            using namespace std;
            int a,b,sum,ii,iii;
            int main(){
               scanf("%d%d",&a,&b);
               for(int i=a;i<=b;i++){
                  ii=i;
                  while(ii!=0){
                     iii=ii%10;
                     if(iii==2){
                        sum++;}
                     ii/=10;}
               }
               printf("%d",sum);
               return 0;}
            
            • 0
              @ 2024-10-14 20:23:15
              #include<bits/stdc++.h>
              using namespace std;
              int l,r,cnt;//l,r:l-r之间的范围
              int sum(int x){//这个函数的意义:判断2出现的次数
              	int cnt=0;
              	while(x>0){//检查各个位数上有没有2
              		if(x%10==2) cnt++;//检查个位
              		x/=10;//去掉个位,再提取个位判断,直到x=0,停止判断
              	}
              	return cnt;//返回cnt
              }
              int main(){
              	cin>>l>>r;
              	for(int i=l;i<=r;i++){
              		cnt+=sum(i);//将这个范围内2出现的次数累加
              	}
              	cout<<cnt;//输出
              	return 0;
              }
              
              • 0
                @ 2023-9-20 0:12:01

                #include<bits/stdc++.h> using namespace std; int main(){ long long l,r,s=0,c,d,a; cin>>l>>r;

                for(int i=l;i<=r;i++){
                c=i;
                for(int j=1;j<=i;j++){
                a=c%10;
                if(a==2){
                s++;
                }
                c=(c-a)/10;
                }
                }
                cout<<s;
                }
                
                • -1
                  @ 2023-9-16 16:50:41
                  
                  

                  #include using namespace std; int main() { int l,r; int s=0; cin>>l>>r; for(int i=l;i<=r;i++) { int q=i; while(q) { if(q%10==2) s++; q/=10; } } cout<<s; }

                  
                  
                  • -1
                    @ 2023-9-16 16:29:39

                    #include using namespace std; int main() { int l,r; int s=0; cin>>l>>r; for(int i=l;i<=r;i++) { int q=i; while(q) { if(q%10==2) s++; q/=10; } } cout<<s; }

                    • 1

                    Information

                    ID
                    488
                    Time
                    1000ms
                    Memory
                    256MiB
                    Difficulty
                    3
                    Tags
                    # Submissions
                    141
                    Accepted
                    76
                    Uploaded By