Chewbacca and Number

#1

Sir/maam
What is the possible error in this program:
Scanner obj = new Scanner(System.in);
long x = obj.nextLong();

	if(x<=Math.pow(10,18)){
		long original = x;
		int inc = 1;
		long rem = 0;
		while(x>0){
			rem = x%10;
			if(rem>4){
				original = original - (rem * inc);
				original = original + ((9 - rem) * inc);
			}
			x/=10;
			inc*=10;
		}
		System.out.println(original);
	}
#2

hi @Ekram… You took long to store the integer while the input constrain is 10^100 … Long can only store number upto range of 10^19 … Please store the number in string and modify your logic

#3

According to the given details:The first line contains a single integer x (1 ≤ x ≤ 10^18)

#4

hi again … actually there might be few cases which can go beyond that value … so it is recommended to try it the string way … if you face a problem there you can send the code after trying …

#5

Scanner obj = new Scanner(System.in); String x = obj.nextLine(); StringBuilder s = new StringBuilder(x); if(s.length()<=Math.pow(10,100)){ char rem = ’ '; for(int i=s.length()-1; i>=0; i–){ rem = s.charAt(i); switch (rem){ case 53: s.setCharAt(i, ‘4’); break; case 54: s.setCharAt(i, ‘3’); break; case 55: s.setCharAt(i, ‘2’); break; case 56: s.setCharAt(i, ‘1’); break; case 57: s.setCharAt(i, ‘0’); break; } } System.out.println(s); }

#6

Hi @Ekram,
please save the code to coding blocks ide and send the URL . I cannot read this code.

#7
#8

Sorry for being so late

#9

hi again ,
See your code works fine for all digits except if there is a 9 at the end or zero at the starting …for example the output of 099999 should be 999999 but your code gives 0 . And there are some changes too as you need to use long to store your number as the input rangee is 1<x<10^18 and the limit of int is only 10^7 . so check this code and optimise it for the above test case .

#10

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