Equal

                            Solution in C :

In C++ :





#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <map>
#include <utility>
#define MOD 1000000007
typedef long long LLD;
using namespace std;
int ar[20000];
int dp[2000];
int main()
{
    int t, n;
    cin >> t;
    while(t--)
    {
	int mn = 1999;
	cin >> n;
	for(int i = 0; i < n; i ++)
	{
	    cin >> ar[i];
	    mn = min(mn, ar[i]);
	}
	for(int i = 0; i < 2000; i ++)
	    dp[i] = MOD;
	dp[0] = 0;
	for(int i = 1; i < 1200; i ++)
	{
	    dp[i] = min(dp[i], dp[i-1] + 1);
	    if(i - 2 >= 0)
		dp[i] = min(dp[i], dp[i-2] + 1);
	    if(i - 5 >= 0)
		dp[i] = min(dp[i], dp[i-5] + 1);
	}
	int steps = MOD;
	for(int s = 0; s <= mn; s ++)
	{
	    int ans = 0;
	    for(int i = 0; i < n; i ++)
		ans += dp[ar[i] - s];

	    steps = min(steps, ans);
	}
	cout << steps << endl;
    }
    return 0;
}









In Java :





import java.io.BufferedInputStream;
import java.util.Scanner;


public class Solution {
	static int offset = 100;
	static Integer f[][] = new Integer[1111][1111];
	static int a[] = new int[11111];
	
	// From i down to j
	static Integer F(int i, int j) {
		if (i < j) return Integer.MAX_VALUE / 2;
		if (f[i + offset][j + offset] != null) return f[i + offset][j + offset];
		if (i == j) return 0;
		int ans = Integer.MAX_VALUE / 2;
		ans = Math.min(ans, F(i - 1, j) + 1);
		ans = Math.min(ans, F(i - 2, j) + 1);
		ans = Math.min(ans, F(i - 5, j) + 1);
		return f[i + offset][j + offset] = ans;
	}
	public static void main(String[] args) {
		Scanner cin = new Scanner(new BufferedInputStream(System.in));
	
		for (int T = cin.nextInt(); T!=0; T--) {
			int n = cin.nextInt();
			int min = Integer.MAX_VALUE;
			for (int i=0; i<n; i++) {
				a[i] = cin.nextInt();
				min = Math.min(min, a[i]);
			}
			int ans = Integer.MAX_VALUE;
			for (int i=min; i>=min-30; i--) {
				int tmp = 0;
				for (int j=0; j<n; j++) {
					tmp += F(a[j], i);
				}
				ans = Math.min(ans, tmp);
			}
			System.out.println(ans);
		}
	}
}









In C :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
	int T;
	scanf("%d\n", &T);
	
	for(int t=0; t<T; t++) {
		int N;
		scanf("%d\n", &N);
		int a[20000];
		
		int min = 0x7fffffff;
		for (int i=0; i<N; i++) {
			scanf("%d", &a[i]);
			if (a[i] < min) min = a[i];
		}
		
		int min_r = 0x7fffffff;
		for( int t=-5; t <= 0; t++) {
			int r = 0;
			for(int i=0; i<N; i++) {
				int diff = a[i] - (min+t);
				r += diff / 5;
				diff = diff % 5;
				r += diff / 2;
				diff = diff % 2;
				r += diff;
			}
			if (r < min_r) min_r = r;
		}
		printf("%d\n", min_r);
	}


    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}









In Python3 :





def steps(number):
    remainder = [0, 1, 1, 2, 2]
    return number // 5 + remainder[number % 5]


T=int(input())
for loop in range(T):
    N = int(input())
    if N==0:
        input()
        print(0)
        continue
    distribution = list(map(int, input().split()))
    distribution.sort()
    minimum = distribution[0]
    totalsteps = [0,1,1,2,2]
    for i in range(5):
        for element in distribution[1:]:
            totalsteps[i] += steps(element-minimum+i)
    print(min(totalsteps))