给定一个正数n。我们需要找到x，以便1 * n，2 * n，3 * n ….. x * n至少给出一次所有10位数字。如果没有，则打印-1。
例子：
Input : n = 1692
Output : 3
Explanation:
n = 1692, we got the digits- 1, 2, 6, 9
2*n = 3384, we got the digits- 1, 2, 3, 4,
6, 8, 9.
3*n = 5076, we got the digits- 1, 2, 3, 4,
5, 6, 7, 8, 9.
At this step we got all the digits at least
once. Therefore our answer is 3.
Input : 1
Output : 10
Input : 0
Output :-1
这里使用的想法很简单。我们从1开始，一直与n相乘，直到至少一次都没有得到所有10位数字。为了跟踪每次迭代中出现的所有数字，我们使用大小为10的临时数组，最初具有全零。每当我们第一次获得一个数字时，我们都会用1初始化它在数组中的索引。当所有数字都被访问一次时，我们就完成了。
以下是它的实现。
C++
// CPP program to find x such that 1*n, 2*n, 3*n
// ...x * n have all digits from 1 to 9 at least
// once
#include
using namespace std;
// Returns smallest value x such that 1*n, 2*n,
// 3*n ...x * n have all digits from 1 to 9 at
// least once
int smallestX(int n)
{
// taking temporary array and variable.
int temp[10] = { 0 };
if (n == 0)
return -1;
// iterate till we get all the 10 digits
// at least once
int count = 0, x = 0;
for (x = 1; count < 10; x++) {
int y = x * n;
// checking all the digits
while (y) {
if (temp[y % 10] == false) {
count++;
temp[y % 10] = true;
}
y /= 10;
}
}
return x - 1;
}
// driver function
int main()
{
int n = 5;
cout < Java
// Java program to find x such
// that 1*n, 2*n, 3*n...x * n
// have all digits from 1 to 9
// at least once
import java.io.*;
import java.util.*;
class GFG
{
// Returns smallest value x
// such that 1*n, 2*n, 3*n
// ...x * n have all digits
// from 1 to 9 at least once
public static int smallestX(int n)
{
// taking temporary
// array and variable.
int[] temp = new int[10];
for(int i = 0; i < 10; i++)
temp[i] = 0;
if (n == 0)
return -1;
// iterate till we get
// all the 10 digits
// at least once
int count = 0, x = 0;
for (x = 1; count < 10; x++)
{
int y = x * n;
// checking all
// the digits
while (y > 0)
{
if (temp[y % 10] == 0)
{
count++;
temp[y % 10] = 1;
}
y /= 10;
}
}
return x - 1;
}
// Driver Code
public static void main(String args[])
{
int n = 5;
System.out.print(smallestX(n));
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)Python3
# Python3 program to find x such
# that 1*n, 2*n, 3*n ...x * n
# have all digits from 1 to 9
# at least once
# Returns smallest value x such
# that 1*n, 2*n, 3*n ...x * n
# have all digits from 1 to 9
# at least once
def smallestX(n):
# taking temporary
# array and variable.
temp = [0]*10
if (n == 0):
return -1
# iterate till we get
# all the 10 digits
# at least once
count = 0
x = 1
while(count < 10):
y = x * n
# checking all
# the digits
while (y>0):
if (temp[y % 10] == 0):
count+=1
temp[y % 10] = 1
y = int(y / 10)
x+=1
return x - 1
# Driver code
if __name__=='__main__':
n = 5
print(smallestX(n))
# This code is contributed
# by mitsC#
// C# program to find x such
// that 1*n, 2*n, 3*n...x * n
// have all digits from 1 to 9
// at least once
using System;
class GFG
{
// Returns smallest value x
// such that 1*n, 2*n, 3*n
// ...x * n have all digits
// from 1 to 9 at least once
public static int smallestX(int n)
{
// taking temporary
// array and variable.
int[] temp = new int[10];
for(int i = 0; i < 10; i++)
temp[i] = 0;
if (n == 0)
return -1;
// iterate till we get
// all the 10 digits
// at least once
int count = 0, x = 0;
for (x = 1; count < 10; x++)
{
int y = x * n;
// checking all the digits
while (y > 0)
{
if (temp[y % 10] == 0)
{
count++;
temp[y % 10] = 1;
}
y /= 10;
}
}
return x - 1;
}
// dDriver Code
static void Main()
{
int n = 5;
Console.Write(smallestX(n));
}
}
// This code is contributed by mitsPHP
输出：
18