求正好能被 100 整除 K 次的第 n 个最小数
原文:https://www . geesforgeks . org/find-n-最小可被 100 整除的数-k 次/
给定两个数 N 和 K，任务是找出第 N 个最小的数，正好除以 100 乘以 K。 例:
输入: N = 12，K = 2 输出: 120000 120000 正好被 100 整除 2 次， 也是第 12 个最小的数。 输入: N = 1000，K = 2 输出: 10010000
进场:
- 首先，找出能被 100 整除 K 次的最小数。也就是 1 之后的 2*K 0，因为 100 只有两个 0。
- 要找到第 N 个最小的数，将 N 乘以我们在加上 2*k 0 后得到的前一个数
- 考虑一个情况，当 N 能被 100 整除，就好像我们把 N 乘以前一个数，那么新的数将有多于(2*k + 1)个尾随的 0，这意味着它能被 100 整除多于 K 倍。
- 将该数字乘以(N + 1)。使用字符串，因为 N 和 K 可能非常大，不适合整数限制。
以下是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the Nth smallest number
string find_number(int N, int K)
{
string r;
// If N is divisible by 100 then we
// multiply N + 1 otherwise, it will be
// divisible by 100 more than K times
if (N % 100 == 0) {
N += 1;
// convert integer to string
r = to_string(N);
}
// if N is not divisible by 100
else {
// convert integer to string
r = to_string(N);
}
// add 2*K 0's at the end to be divisible
// by 100 exactly K times
for (int i = 1; i <= K; i++)
r += "00";
return r;
}
// Driver Code
int main()
{
int N = 1000, K = 2;
string ans = find_number(N, K);
cout << ans << "\n";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of above approach
import java.util.*;
class GFG
{
// Function to find the Nth smallest number
static String find_number(int N, int K)
{
String r;
// If N is divisible by 100 then we
// multiply N + 1 otherwise, it will be
// divisible by 100 more than K times
if (N % 100 == 0)
{
N += 1;
// convert integer to string
r = String.valueOf(N);
}
// if N is not divisible by 100
else
{
// convert integer to string
r = String.valueOf(N);
}
// add 2*K 0's at the end to be divisible
// by 100 exactly K times
for (int i = 1; i <= K; i++)
r += "00";
return r;
}
// Driver Code
public static void main(String[] args)
{
int N = 1000, K = 2;
String ans = find_number(N, K);
System.out.println(ans);
}
}
/* This code is contributed by PrinciRaj1992 */
Python 3
# Python3 implementation of above approach
# Function to find the Nth smallest number
def find_number(N, K):
r = ""
# If N is divisible by 100 then we
# multiply N + 1 otherwise, it will be
# divisible by 100 more than K times
if (N % 100 == 0):
N += 1;
# convert integer to string
r = str(N)
# if N is not divisible by 100
else:
# convert integer to string
r = str(N)
# add 2*K 0's at the end to be divisible
# by 100 exactly K times
for i in range(1, K + 1):
r += "00"
return r
# Driver Code
N = 1000
K = 2;
ans = find_number(N, K)
print(ans)
# This code is contributed by Mohit Kumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to find the Nth smallest number
static String find_number(int N, int K)
{
String r;
// If N is divisible by 100 then we
// multiply N + 1 otherwise, it will be
// divisible by 100 more than K times
if (N % 100 == 0)
{
N += 1;
// convert integer to string
r = N.ToString();
}
// if N is not divisible by 100
else
{
// convert integer to string
r = N.ToString();
}
// add 2*K 0's at the end to be divisible
// by 100 exactly K times
for (int i = 1; i <= K; i++)
r += "00";
return r;
}
// Driver Code
public static void Main(String[] args)
{
int N = 1000, K = 2;
String ans = find_number(N, K);
Console.WriteLine(ans);
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// JavaScript implementation of above approach
// Function to find the Nth smallest number
function find_number(N, K)
{
var r;
// If N is divisible by 100 then we
// multiply N + 1 otherwise, it will be
// divisible by 100 more than K times
if (N % 100 == 0)
{
N += 1;
// convert integer to string
r = N.toString();
}
// if N is not divisible by 100
else
{
// convert integer to string
r = N.toString();
}
// add 2*K 0's at the end to be divisible
// by 100 exactly K times
for(var i = 1; i <= K; i++)
r += "00";
return r;
}
// Driver Code
var N = 1000, K = 2;
var ans = find_number(N, K);
document.write(ans);
// This code is contributed by Khushboogoyal499
</script>
Output:
10010000
时间复杂度: O(K)
版权属于：月萌API www.moonapi.com，转载请注明出处
