DivToZero
SRM 262 · 2005-09-09 · by Softwalker
SRM 262 · 2005-09-09 · by Softwalker · Simple Math
Problem Statement
Problem Statement
You are given an integer num from which you should replace the last two digits such that the resulting number is divisible by factor and is also the smallest possible number. Return the two replacement digits as a String .
For instance:
if num = 275, and factor = 5, you would return "00" because 200 is divisible by 5.
if num = 1021, and factor = 11, you would return "01" because 1001 is divisible by 11.
if num = 70000, and factor = 17, you would return "06" because 70006 is divisible by 17.
For instance:
if num = 275, and factor = 5, you would return "00" because 200 is divisible by 5.
if num = 1021, and factor = 11, you would return "01" because 1001 is divisible by 11.
if num = 70000, and factor = 17, you would return "06" because 70006 is divisible by 17.
Constraints
- factor must be between 1 and 100, inclusive.
- num must be between 100 and 2,000,000,000, inclusive.
Examples
0)
2000000000 100 Returns: "00"
1)
1000 3 Returns: "02"
2)
23442 75 Returns: "00"
3)
428392 17 Returns: "15"
4)
32442 99 Returns: "72"
Submissions are judged against all 44 archived test cases, of which 5 are shown here. Case numbers match the judge’s.
Coding Area
Language: C++17 · define a public class DivToZero with a public method string lastTwo(int num, int factor) · 44 test cases · 2 s / 256 MB per case