给定两个正整数N和K ，任务是找到数字总和为N的最小数字，并且该数字中的每个不同数字最多出现K次。如果不存在这样的数字，则打印“ -1” 。
例子：
Input: N = 25, K = 3
Output: 799
Explanation: Sum of digits of the number = (7 + 9 + 9) = 25 and is the smallest possible.
Input: N =100, K = 2
Output: -1
方法：可以使用贪婪方法解决给定的问题。请按照以下步骤解决问题：
- 每个数字最多出现K次的任何数字的最大可能数字总和为K * 45 。
- 初始化一个字符串，例如res ，以存储形成的结果字符串。
- 如果N大于K * 45 ，则无法获得该数字。因此，打印“ -1” 。
- 否则，请最多从K减去从N到9到1的每个数字。将当前数字添加到res 。继续直到N小于当前数字。
- 如果N小于当前数字，则将N作为数字插入res 。
- 完成上述步骤后，以相反的顺序打印字符串res作为要求的答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the smallest number
// whose sum of digits is N and each
// digit occurring at most K times
void findSmallestNumberPossible(int N, int K)
{
// Maximum sum possible using
// each digit at most K times
if (N > 45 * K) {
cout << "-1";
return;
}
// Append smallest number into
// the resultant string
string res = "";
int count = 0;
// Iterate over the range [9, 1]
for (int i = 9; i >= 1;) {
// If the count of the digit
// is K, then update count and
// check for the next digit
if (count == K) {
i--;
count = 0;
}
// If N is greater than
// current digit
if (N > i) {
// Subtract i from N
N -= i;
// Insert i as a digit
res += (char)48 + i;
}
else {
// Insert remaining N
// as a digit
res += (char)N + 48;
N = 0;
break;
}
// Increment count
count++;
}
// Reverse the string
reverse(res.begin(),
res.end());
// Print the answer
cout << res;
}
// Driver Code
int main()
{
int N = 25, K = 3;
// Function Call
findSmallestNumberPossible(N, K);
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG{
// Function to find the smallest number
// whose sum of digits is N and each
// digit occurring at most K times
static void findSmallestNumberPossible(int N, int K)
{
// Maximum sum possible using
// each digit at most K times
if (N > 45 * K)
{
System.out.print("-1");
return;
}
// Append smallest number into
// the resultant string
StringBuilder res = new StringBuilder();
int count = 0;
// Iterate over the range [9, 1]
for(int i = 9; i >= 1😉
{
// If the count of the digit
// is K, then update count and
// check for the next digit
if (count == K)
{
i--;
count = 0;
}
// If N is greater than
// current digit
if (N > i)
{
// Subtract i from N
N -= i;
// Insert i as a digit
res.append((char)('0' + i));
}
else
{
// Insert remaining N
// as a digit
res.append((char)('0' + N));
N = 0;
break;
}
// Increment count
count++;
}
// Reverse the string
res.reverse();
// Print the answer
System.out.print(res.toString());
}
// Driver Code
public static void main(String[] args)
{
int N = 25, K = 3;
// Function Call
findSmallestNumberPossible(N, K);
}
}
// This code is contributed by jithinPython3
# Python3 program for the above approach
# Function to find the smallest number
# whose sum of digits is N and each
# digit occurring at most K times
def findSmallestNumberPossible(N, K) :
# Maximum sum possible using
# each digit at most K times
if (N > 45 * K) :
print("-1", endl = "");
return;
# Append smallest number into
# the resultant string
res = "";
count = 0;
# Iterate over the range [9, 1]
i = 9;
while i >= 1 :
# If the count of the digit
# is K, then update count and
# check for the next digit
if (count == K) :
i -= 1;
count = 0;
# If N is greater than
# current digit
if (N > i) :
# Subtract i from N
N -= i;
# Insert i as a digit
res += chr(48 + i);
else :
# Insert remaining N
# as a digit
res += chr(N + 48);
N = 0;
break;
# Increment count
count += 1;
# Reverse the string
res = res[::-1]
# Print the answer
print(res, end = "");
# Driver Code
if __name__ == "__main__" :
N = 25; K = 3;
# Function Call
findSmallestNumberPossible(N, K);
# This code is contributed by AnkThonC#
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to find the smallest number
// whose sum of digits is N and each
// digit occurring at most K times
static void findSmallestNumberPossible(int N, int K)
{
// Maximum sum possible using
// each digit at most K times
if (N > 45 * K)
{
Console.Write("-1");
return;
}
// Append smallest number into
// the resultant string
string res = "" ;
int count = 0;
// Iterate over the range [9, 1]
for (int i = 9; i >= 1;)
{
// If the count of the digit
// is K, then update count and
// check for the next digit
if (count == K)
{
i--;
count = 0;
}
// If N is greater than
// current digit
if (N > i)
{
// Subtract i from N
N -= i;
// Insert i as a digit
res += (char)('0' + i);
}
else
{
// Insert remaining N
// as a digit
res += (char)('0' + N);
N = 0;
break;
}
// Increment count
count++;
}
// Reverse the string
string ans = "";
for (int i = res.Length - 1; i >= 0; i--)
{
ans += res[i] + "";
}
// Print the answer
Console.WriteLine(ans);
}
// Driver Code
public static void Main()
{
int N = 25, K = 3;
// Function Call
findSmallestNumberPossible(N, K);
}
}
// This code is contributed by susmitakundugoaldanga输出：
799时间复杂度： O(log 10 N)
辅助空间： O(1)