1 solutions

  • 0
    @ 2025-3-3 16:34:00

    C :

    #include<stdio.h>   
    
    int main(){ 
    	int i,n,c=0;
    	scanf("%d",&n);
    
    	
    	for(i = 5;i <= n;i++){
    		int w = i / 10000;
    		int q = i / 1000 % 10;
    		int b = i / 100 % 10;
    		int s = i / 10 % 10;
    		int g = i % 10;
    		
    		if((w == 5 || q == 5 || b == 5 || s == 5 || g == 5) && i % 5 == 0){
    		
    			c++;
    		}
    	}
    	
    	printf("%d",c);
    	return 0;
    } 
    

    C++ :

    #include<iostream>
    #include<math.h>
    using namespace std;
    int main(){
        int n,i,a,b,c,d,e,f = 0;
        cin>>n;
        i = 1;
        while(i <= n){ 
      a = i / 10000;
      b = i / 1000 % 10;
      c = i / 100 % 10;
      d = i / 10 % 10;
      e = i % 10;
      if((i % 5 == 0) && (a == 5 || b == 5 || c == 5 || d == 5 || e == 5) ){
          f++;
      }
      i++;
    }
    cout<<f<<endl;
    return 0;
    }
    
    
    • 1

    【入门】能被5整除且至少有一位数字是5的所有整数的个数

    Information

    ID
    10892
    Time
    1000ms
    Memory
    16MiB
    Difficulty
    (None)
    Tags
    # Submissions
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