给定一个字符串str和一个整数K ，任务是找到一个字符串S ，使得它具有给定字符串str的正好K个子序列。
例子：
Input: str = “gfg”, K = 10
Output: gggggffg
Explanation:
There are 10 possible subsequence of the given string “gggggffg”. They are:
1. gggggffg
2. gggggffg
3. gggggffg
4. gggggffg
5. gggggffg
6. gggggffg
7. gggggffg
8. gggggffg
9. gggggffg
10. gggggffg.
Input: str = “code”, K = 20
Output: cccccoodde
Explanation:
There are 20 possible subsequence of the string “cccccoodde”.
方法：
为了解决上述问题，我们必须遵循以下步骤：
- 这个想法是找到K的素因数并存储素因数(比如说factor )。
- 创建给定字符串大小的空数组计数，以将每个字符的计数存储在结果字符串s中。用1初始化数组。
- 现在，从列表因子弹出元素，并以循环方式将其乘以数组的每个位置，直到列表变空。最后，我们获得了数组中str的每个字符的数量。
- 迭代数组count [] ，并将每个字符ch的字符数附加到结果字符串s上。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
#include
using namespace std;
// Function that computes the string s
void printSubsequenceString(string str,
long long k)
{
// Length of the given string str
int n = str.size();
int i;
// List that stores all the prime
// factors of given k
vector factors;
// Find the prime factors
for (long long i = 2;
i <= sqrt(k); i++) {
while (k % i == 0) {
factors.push_back(i);
k /= i;
}
}
if (k > 1)
factors.push_back(k);
// Initialize the count of each
// character position as 1
vector count(n, 1);
int index = 0;
// Loop until the list
// becomes empty
while (factors.size() > 0) {
// Increase the character
// count by multiplying it
// with the prime factor
count[index++] *= factors.back();
factors.pop_back();
// If we reach end then again
// start from beginning
if (index == n)
index = 0;
}
// Store the output
string s;
for (i = 0; i < n; i++) {
while (count[i]-- > 0) {
s += str[i];
}
}
// Print the string
cout << s;
}
// Driver code
int main()
{
// Given String
string str = "code";
long long k = 20;
// Function Call
printSubsequenceString(str, k);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function that computes the String s
static void printSubsequenceString(String str,
int k)
{
// Length of the given String str
int n = str.length();
int i;
// List that stores all the prime
// factors of given k
Vector factors = new Vector();
// Find the prime factors
for (i = 2; i <= Math.sqrt(k); i++)
{
while (k % i == 0)
{
factors.add(i);
k /= i;
}
}
if (k > 1)
factors.add(k);
// Initialize the count of each
// character position as 1
int []count = new int[n];
Arrays.fill(count, 1);
int index = 0;
// Loop until the list
// becomes empty
while (factors.size() > 0)
{
// Increase the character
// count by multiplying it
// with the prime factor
count[index++] *= factors.get(factors.size() - 1);
factors.remove(factors.get(factors.size() - 1));
// If we reach end then again
// start from beginning
if (index == n)
index = 0;
}
// Store the output
String s = "";
for (i = 0; i < n; i++)
{
while (count[i]-- > 0)
{
s += str.charAt(i);
}
}
// Print the String
System.out.print(s);
}
// Driver code
public static void main(String[] args)
{
// Given String
String str = "code";
int k = 20;
// Function Call
printSubsequenceString(str, k);
}
}
// This code is contributed by sapnasingh4991 Python3
# Python3 program for
# the above approach
import math
# Function that computes
# the string s
def printSubsequenceString(st, k):
# Length of the given
# string str
n = len(st)
# List that stores
# all the prime
# factors of given k
factors = []
# Find the prime factors
sqt = (int(math.sqrt(k)))
for i in range (2, sqt + 1):
while (k % i == 0):
factors.append(i)
k //= i
if (k > 1):
factors.append(k)
# Initialize the count of each
# character position as 1
count = [1] * n
index = 0
# Loop until the list
# becomes empty
while (len(factors) > 0):
# Increase the character
# count by multiplying it
# with the prime factor
count[index] *= factors[-1]
factors.pop()
index += 1
# If we reach end then again
# start from beginning
if (index == n):
index = 0
# store output
s = ""
for i in range (n):
while (count[i] > 0):
s += st[i]
count[i] -= 1
# Print the string
print (s)
# Driver code
if __name__ == "__main__":
# Given String
st = "code"
k = 20
# Function Call
printSubsequenceString(st, k)
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function that computes the String s
static void printSubsequenceString(String str,
int k)
{
// Length of the given String str
int n = str.Length;
int i;
// List that stores all the prime
// factors of given k
List factors = new List();
// Find the prime factors
for (i = 2; i <= Math.Sqrt(k); i++)
{
while (k % i == 0)
{
factors.Add(i);
k /= i;
}
}
if (k > 1)
factors.Add(k);
// Initialize the count of each
// character position as 1
int []count = new int[n];
for (i = 0; i < n; i++)
count[i] = 1;
int index = 0;
// Loop until the list
// becomes empty
while (factors.Count > 0)
{
// Increase the character
// count by multiplying it
// with the prime factor
count[index++] *= factors[factors.Count - 1];
factors.Remove(factors[factors.Count - 1]);
// If we reach end then again
// start from beginning
if (index == n)
index = 0;
}
// Store the output
String s = "";
for (i = 0; i < n; i++)
{
while (count[i]-- > 0)
{
s += str[i];
}
}
// Print the String
Console.Write(s);
}
// Driver code
public static void Main(String[] args)
{
// Given String
String str = "code";
int k = 20;
// Function Call
printSubsequenceString(str, k);
}
}
// This code is contributed by sapnasingh4991 输出：
cccccoodde
时间复杂度： O(N * log 2 (log 2 (N)))
辅助空间： O(K)