B. Ordering Number
Time Limit: 1.0 s
Memory Limit: 64.0 MB
Number v is a Divisor of a number u, if u is divisible by v.
Given an array, order it by the following condition.
x will come before y if
- number of divisors of x is less than number of divisors of y
- number of divisors of x is equal to number of divisors of y and x > y.
After ordering, print the kth number.
Input format:
T number of case, T <= 100
N = size of array, N <= 1000
A[i].....A[n] = array. A[i] <= 10000
K = kth number to print. K <= N
Output format:
X, where X = kth number after ordering the array.
Sample Input:
2
10
1 2 3 4 5 6 7 8 9 10
5
5
1 2 3 5 7
3
Sample Output:
2
5
Hint
C Code
#include <stdio.h>
int main(void)
{
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", a + b);
return 0;
}
C++ Code
#include <iostream>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
cout << a + b << endl;
return 0;
}
C# Code
using System;
using System.Linq;
class Program {
public static void Main(string[] args) {
Console.WriteLine(Console.ReadLine().Split().Select(int.Parse).Sum());
}
}
Python3 Code
a, b = list(map(int, input().split()))
print(a + b)
Java Code
import java.io.*;
import java.util.Scanner;
public class Main {
/**
* @param args
* @throws IOException
*/
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
int b = sc.nextInt();
System.out.println(a + b);
}
}
Ruby Code
a, b = gets.split.map(&:to_i)
puts(a + b)
Source
Information
- ID
- 1008
- Difficulty
- 3
- Category
- Implementation | Sorting Click to Show
- Tags
- # Submissions
- 31
- Accepted
- 19
- Accepted Ratio
- 61%
- Uploaded By
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