Algorithm
Problem Name: Mathematics -
https://www.hackerrank.com/challenges/constructing-a-number/problem?isFullScreen=true
In this HackerRank in Mathematics -
Manipulating numbers is at the core of a programmer's job. To test how well you know their properties, you are asked to solve the following problem.
You are given n non-negative integers a1,a2, .... , an . You want to know whether it's possible to construct a new integer using all the digits of these numbers such that it would be divisible by 3 . You can reorder the digits as you want. The resulting number can contain leading zeros.
For example, consider the numbers 50,40,90 from which you have to construct a new integer as described above. Numerous arrangements of digits are possible; but we have illustrated one below.
Complete the function canConstruct which takes an integer array as input and return "Yes" or "No" based on whether or not the required integer can be formed.
Input Format
The first line contains a single integer t denoting the number of queries. The following lines describe the queries. Each query is described in two lines. The first of these lines contains a single integer n. The second contains n space-separated integers a1,a2, .... , an .
Constraints
- 1 <= t <= 1000
- 1 <= n <= 1000
- 1 <= ai <= 103
Subtasks
For 33.33% of the total score:
- n = 1
- 1 <= a1 <= 106
Output Format
For each query, print a single line containing "Yes" if it's possible to construct such integer and "No" otherwise.
Sample Input 0
3
1
9
3
40 50 90
2
1 4
Sample Output 0
Yes
Yes
No
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main()
{
unsigned long long int r,c;
scanf("%llu%llu",&r,&c);
if(r%2 == 0)
{
unsigned long long int term = ((r-1)/2)*5+c;
printf("%llu",2*term-1); // nth term in odd series = 2*n-1
}
else{
unsigned long long int term = ((r)/2)*5+c;
printf("%llu", 2*term-2); // nth term in even series = 2*n-2
}
return 0;
}
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <cstdio>
using namespace std;
int main() {
while(getchar_unlocked() > '\n')
{}
int c = 0;
while(c >= 0) {
while(getchar_unlocked() > '\n')
{}
int n = 0;
while((c = getchar_unlocked()) > '\n') {
if(c >= '0')
n += c;
}
puts(n % 3 ? "No" : "Yes");
}
return 0;
}