给定两个数字N和M ，任务是查找由M组成的两个数字作为其所有数字，以使它们的差可被N整除。
例子：
Input: N = 8, M = 2
Output: 22 222
Explanation: The difference between 222 and 22 (200) is divisible by 8
Input: N = 17, M = 6
Output: 6 66666666666666666
方法：
在此问题中，我们必须找到仅包含一个唯一数字的数字。假设M等于2，那么我们必须从2、22、222、2222之类的数字中找到A和B……依此类推。 A和B之间的差异应被N整除。为了满足此条件，我们必须选择A和B，以使A和B的其余部分除以N相同。
对于长度为N + 1长度的数字(仅包含一个唯一的数字M) ，我们将有N + 1个数字。如果我们将这些N + 1个数字除以N，我们将有N + 1个余数，范围为[0，N] 。由于数字可以超出整数值的范围，因此我们将数字的剩余长度作为键值对存储在Map中。一旦剩余值与Map中已经配对的值一起出现，则当前长度和映射的长度就是所需数字的长度。
下面的代码是上述方法的实现：
C++
// C++ implementation
// of the above approach
#include
using namespace std;
// Function to implement
// the above approach
void findNumbers(int N, int M)
{
int m = M;
// Hashmap to store
// remainder-length of the
// number as key-value pairs
map remLen;
int len, remainder;
// Iterate till N + 1 length
for (len = 1; len <= N + 1; ++len) {
remainder = M % N;
// Search remainder in the map
if (remLen.find(remainder)
== remLen.end())
// If remainder is not
// already present insert
// the length for the
// corresponding remainder
remLen[remainder] = len;
else
break;
// Keep increasing M
M = M * 10 + m;
// To keep M in range of integer
M = M % N;
}
// Length of one number
// is the current Length
int LenA = len;
// Length of the other number
// is the length paired with
// current remainder in map
int LenB = remLen[remainder];
for (int i = 0; i < LenB; ++i)
cout << m;
cout << " ";
for (int i = 0; i < LenA; ++i)
cout << m;
return;
}
// Driver code
int main()
{
int N = 8, M = 2;
findNumbers(N, M);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class GFG{
// Function to implement
// the above approach
static void findNumbers(int N, int M)
{
int m = M;
// Hashmap to store
// remainder-length of the
// number as key-value pairs
Map remLen = new HashMap<>();
int len, remainder = 0;
// Iterate till N + 1 length
for(len = 1; len <= N + 1; ++len)
{
remainder = M % N;
// Search remainder in the map
if (!remLen.containsKey(remainder))
{
// If remainder is not
// already present insert
// the length for the
// corresponding remainder
remLen.put(remainder, len);
}
else
{
break;
}
// Keep increasing M
M = M * 10 + m;
// To keep M in range of integer
M = M % N;
}
// Length of one number
// is the current Length
int LenA = len;
// Length of the other number
// is the length paired with
// current remainder in map
int LenB = remLen.getOrDefault(remainder, 0);
for(int i = 0; i < LenB; ++i)
System.out.print(m);
System.out.print(" ");
for(int i = 0; i < LenA; ++i)
System.out.print(m);
}
// Driver code
public static void main(String[] args)
{
int N = 8, M = 2;
findNumbers(N, M);
}
}
// This code is contributed by offbeat Python3
# Python3 implementation
# of the above approach
# Function to implement
# the above approach
def findNumbers(N, M):
m = M
# Hashmap to store
# remainder-length of the
# number as key-value pairs
remLen = {}
# Iterate till N + 1 length
for len1 in range(1, N + 1, 1):
remainder = M % N
# Search remainder in the map
if (remLen.get(remainder) == None):
# If remainder is not
# already present insert
# the length for the
# corresponding remainder
remLen[remainder] = len1
else:
break
# Keep increasing M
M = M * 10 + m
# To keep M in range of integer
M = M % N
# Length of one number
# is the current Length
LenA = len1
# Length of the other number
# is the length paired with
# current remainder in map
LenB = remLen[remainder]
for i in range(LenB):
print(m, end = "")
print(" ", end = "")
for i in range(LenA):
print(m, end = "")
return
# Driver code
if __name__ == '__main__':
N = 8
M = 2
findNumbers(N, M)
# This code is contributed by Bhupendra_SinghC#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to implement
// the above approach
static void findNumbers(int N, int M)
{
int m = M;
// To store remainder-length of
// the number as key-value pairs
Dictionary remLen = new Dictionary();
int len, remainder = 0;
// Iterate till N + 1 length
for(len = 1; len <= N + 1; ++len)
{
remainder = M % N;
// Search remainder in the map
if (!remLen.ContainsKey(remainder))
{
// If remainder is not
// already present insert
// the length for the
// corresponding remainder
remLen.Add(remainder, len);
}
else
{
break;
}
// Keep increasing M
M = M * 10 + m;
// To keep M in range of integer
M = M % N;
}
// Length of one number
// is the current Length
int LenA = len;
// Length of the other number
// is the length paired with
// current remainder in map
int LenB = remLen[remainder];
for(int i = 0; i < LenB; ++i)
Console.Write(m);
Console.Write(" ");
for(int i = 0; i < LenA; ++i)
Console.Write(m);
}
// Driver code
public static void Main(String[] args)
{
int N = 8, M = 2;
findNumbers(N, M);
}
}
// This code is contributed by Amit Katiyar 输出：
22 222
时间复杂度： O(N)
辅助空间： O(1)
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