C语言解答

#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++解答

#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解答

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 ;
   }
}