给定一个由N 个0或7整数组成的数组arr[] ，任务是找到可以使用数组元素形成的最大数，使得它可以被50整除。
例子：
Input: arr[] = {7, 7, 7, 7, 7, 7, 0, 0, 0, 0, 0, 0, 0}
Output: 777770000000
Input: arr[] = {7, 0}
Output: 0
朴素的方法：解决这个问题的最简单的方法是基于以下观察：
- 可视化50等于5 * 10 ，因此插入的任何尾随零将除以10作为因子50 。因此，任务简化为组合7秒，以便它们可以被5整除，并且对于可以被5整除的数字，其单位位置应该是0或5 。
时间复杂度： O(2 N )
辅助空间： O(1)
高效方法：为了优化上述方法，其思想是计算数组中7 s 和0 s 的频率，并生成可被50整除的所需数字。请按照以下步骤解决问题：
- 计算数组中存在的7秒和0秒的数量。
- 计算与数组中存在的7 s 计数最接近的因子 5 （因为35是仅使用 7s 可以获得的最小因子5 ）
- 显示计算出的7个数。
- 将尾随零附加到上面所做的数字。
需要考虑的角落案例：
- If the count of 0s the array is 0(then, the task is to check if the count of 7s in the array is divisible by 50 or not.
- If the count of 7s is less than 5 and no 0 is present in the array, simply print “Not Possible”.
- If the count of 7s is less than 5 and 0 is present, simply print ‘0‘.
下面是上述方法的实现：
C++
// C++ Program for the above approach
#include
using namespace std;
// Print the largest number divisible by 50
void printLargestDivisible(int arr[], int N)
{
int i, count0 = 0, count7 = 0;
for (i = 0; i < N; i++) {
// Counting number of 0s and 7s
if (arr[i] == 0)
count0++;
else
count7++;
}
// If count of 7 is divisible by 50
if (count7 % 50 == 0) {
while (count7--)
cout << 7;
while (count0--)
cout << 0;
}
// If count of 7 is less than 5
else if (count7 < 5) {
if (count0 == 0)
cout << "No";
else
cout << "0";
}
// If count of 7 is not
// divisible by 50
else {
// Count of groups of 5 in which
// count of 7s can be grouped
count7 = count7 - count7 % 5;
while (count7--)
cout << 7;
while (count0--)
cout << 0;
}
}
// Driver Code
int main()
{
// Given array
int arr[] = { 0, 7, 0, 7, 7, 7, 7, 0,
0, 0, 0, 0, 0, 7, 7, 7 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
printLargestDivisible(arr, N);
return 0;
} Java
// Java program of the above approach
import java.io.*;
class GFG {
// Print the largest number divisible by 50
static void printLargestDivisible(int arr[], int N)
{
int i, count0 = 0, count7 = 0;
for (i = 0; i < N; i++) {
// Counting number of 0s and 7s
if (arr[i] == 0)
count0++;
else
count7++;
}
// If count of 7 is divisible by 50
if (count7 % 50 == 0) {
while (count7 != 0)
{
System.out.print(7);
count7 -= 1;
}
while (count0 != 0)
{
System.out.print(0);
count0 -= 1;
}
}
// If count of 7 is less than 5
else if (count7 < 5) {
if (count0 == 0)
System.out.print("No");
else
System.out.print( "0");
}
// If count of 7 is not
// divisible by 50
else {
// Count of groups of 5 in which
// count of 7s can be grouped
count7 = count7 - count7 % 5;
while (count7 != 0)
{
System.out.print(7);
count7 -= 1;
}
while (count0 != 0)
{
System.out.print(0);
count0 -= 1;
}
}
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 0, 7, 0, 7, 7, 7, 7, 0,
0, 0, 0, 0, 0, 7, 7, 7 };
// Size of the array
int N = arr.length;
printLargestDivisible(arr, N);
}
}
// This code is contributed by jana_sayantan.Python3
# Python3 Program for the above approach
# Print the largest number divisible by 50
def printLargestDivisible(arr, N) :
count0 = 0; count7 = 0;
for i in range(N) :
# Counting number of 0s and 7s
if (arr[i] == 0) :
count0 += 1;
else :
count7 += 1;
# If count of 7 is divisible by 50
if (count7 % 50 == 0) :
while (count7) :
count7 -= 1;
print(7, end = "");
while (count0) :
count0 -= 1;
print(count0, end = "");
# If count of 7 is less than 5
elif (count7 < 5) :
if (count0 == 0) :
print("No", end = "");
else :
print("0", end = "");
# If count of 7 is not
# divisible by 50
else :
# Count of groups of 5 in which
# count of 7s can be grouped
count7 = count7 - count7 % 5;
while (count7) :
count7 -= 1;
print(7, end = "");
while (count0) :
count0 -= 1;
print(0, end = "");
# Driver Code
if __name__ == "__main__" :
# Given array
arr = [ 0, 7, 0, 7, 7, 7, 7, 0, 0, 0, 0, 0, 0, 7, 7, 7 ];
# Size of the array
N = len(arr);
printLargestDivisible(arr, N);
# This code is contributed by AnkThonC#
// C# program of the above approach
using System;
public class GFG {
// Print the largest number divisible by 50
static void printLargestDivisible(int []arr, int N)
{
int i, count0 = 0, count7 = 0;
for (i = 0; i < N; i++)
{
// Counting number of 0s and 7s
if (arr[i] == 0)
count0++;
else
count7++;
}
// If count of 7 is divisible by 50
if (count7 % 50 == 0) {
while (count7 != 0)
{
Console.Write(7);
count7 -= 1;
}
while (count0 != 0)
{
Console.Write(0);
count0 -= 1;
}
}
// If count of 7 is less than 5
else if (count7 < 5) {
if (count0 == 0)
Console.Write("No");
else
Console.Write( "0");
}
// If count of 7 is not
// divisible by 50
else {
// Count of groups of 5 in which
// count of 7s can be grouped
count7 = count7 - count7 % 5;
while (count7 != 0)
{
Console.Write(7);
count7 -= 1;
}
while (count0 != 0)
{
Console.Write(0);
count0 -= 1;
}
}
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int []arr = { 0, 7, 0, 7, 7, 7, 7, 0,
0, 0, 0, 0, 0, 7, 7, 7 };
// Size of the array
int N = arr.Length;
printLargestDivisible(arr, N);
}
}
// This code is contributed by shikhasingrajputJavascript
输出：
7777700000000时间复杂度： O(N)
辅助空间： O(1)