1 条题解

  • 0
    @ 2025-4-7 21:19:28

    C :

    #include<stdio.h>
    #include<string.h>
    int maxsize=20;
    int calc(char a[], char b[]){
      int lena, lenb, i, j, result;
      lena=strlen(a);
      lenb=strlen(b);
      result=0;
      for(i=0; i<lena; i++){
        for(j=0; j<lenb; j++){
          result+=(a[i]-'0')*(b[j]-'0');
        }
      }
      return result;
    }
      
    int main(){
      char a[maxsize], b[maxsize];
      int result;
      while(scanf("%s%s", a, b)==2){
        result=calc(a, b);
        printf("%d\n", result);
      }
      return 0;
    }
    

    C++ :

    #include <stdio.h>
    int a,b;
    int run()
    {
    	int i=0,j=0;
    	while(a!=0)
    	{
    		i+=a%10;
    		a/=10;
    	}
    	while(b!=0)
    	{
    		j+=b%10;
    		b/=10;
    	}
    	printf("%d\n",i*j);
    }
    int main()
    {
    	scanf("%d%d",&a,&b);
    	while((a!=-9999999)||(b!=-4444444))
    	{
    		run();
    		a=-9999999;
    		b=-4444444;
    		scanf("%d%d",&a,&b);
    	}
    	return 0;
    }
    

    Java :

    
    
    import java.util.Scanner;
    
    public class Main {
       private static Scanner s = new Scanner(System.in) ;
       
       public static void main(String[] args) {
    	   while(s.hasNext()){
    		   long a = s.nextLong() ;
    		   long b = s.nextLong() ;
    		   System.out.println(f(a, b)) ;
    	   }
    	   
    	   
       }
       
       public static long f(long a , long b){
    	   long x = a ;
    	   long y = b ;
    	   int i = 0 ;
    	   int j = 0 ;
    	   while(a>0){
    		 a = a/10 ;
    		 i++ ;
    	   }
    	   while(b>0){
    			 b = b/10 ;
    			 j++ ;
    		   }
    		   
    	   int c [] = new int[i] ;
    	   int d [] = new int[j] ;
    	   i = 0 ;
    	   j = 0 ;
    	   while(x>0){
    		   c[i] = (int) (x%10) ;
    		   i++ ;
    		   x = x /10 ;
    	   }
    	   while(y>0){
    		   d[j] = (int) (y%10) ;
    		   j++ ;
    		   y = y /10 ;
    	   }
    	   long sum = 0 ;
    	   for (int k = 0; k < c.length; k++) {
    		for (int k2 = 0; k2 < d.length; k2++) {
    			sum = sum+c[k]*d[k2] ; 
    		}
    	   }
    	   
    	   return sum ;
       }
    }
    
    
    • 1

    信息

    ID
    1106
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者