找出能被 D 整除的 N 位数
原文:https://www . geesforgeks . org/find-n-digits-number-可被 d 整除/
给定
和
。任务是找到一个能被 D 整除的 N 位数(2 < = D < = 10)。如果不可能，则打印不可能。
示例:
Input : N = 2 and D = 2
Output : 20
Input : N = 1 and D = 10
Output : Impossible
进场:有两个条件，D=10，D 不等于 10。如果 D = 10，N = 1，那么唯一的答案是不可能的，在所有其他条件下，答案都是可能的。
1\. If D is 10,
Print 1 followed by n-1 times zero.
2\. If D is not 10
Print D followed by n-1 times zero
下面是上述方法的实现:
C++
// CPP program to Find N digits
// number which is divisible by D
#include <bits/stdc++.h>
using namespace std;
// Function to return N digits
// number which is divisible by D
string findNumber(int n, int d)
{
// to store answer
string ans = "";
if (d != 10) {
ans += to_string(d);
for (int i = 1; i < n; i++)
ans += '0';
}
else {
if (n == 1)
ans += "Impossible";
else {
ans += '1';
for (int i = 1; i < n; i++)
ans += '0';
}
}
return ans;
}
// Driver code
int main()
{
int n = 12, d = 3;
cout << findNumber(n, d);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to Find N digits
// number which is divisible by D
import java.io.*;
class GFG {
// Function to return N digits
// number which is divisible by D
static String findNumber(int n, int d)
{
// to store answer
String ans = "";
if (d != 10) {
ans += Integer.toString(d);
for (int i = 1; i < n; i++)
ans += '0';
}
else {
if (n == 1)
ans += "Impossible";
else {
ans += '1';
for (int i = 1; i < n; i++)
ans += '0';
}
}
return ans;
}
// Driver code
public static void main (String[] args) {
int n = 12, d = 3;
System.out.println(findNumber(n, d));
}
}
// This code is contributed by anuj_67..
Python 3
# Python 3 program to Find N digits
# number which is divisible by D
# Function to return N digits
# number which is divisible by D
def findNumber(n, d):
# to store answer
ans = ""
if (d != 10) :
ans += str(d)
for i in range(1,n):
ans += '0'
else :
if (n == 1):
ans += "Impossible"
else :
ans += '1'
for i in range(1,n):
ans += '0'
return ans
# Driver code
if __name__ == "__main__":
n = 12
d = 3
print(findNumber(n, d))
# This code is contributed by
# ChitraNayal
C
// C# program to Find N digits
// number which is divisible by D
using System;
class GFG {
// Function to return N digits
// number which is divisible by D
static string findNumber(int n, int d)
{
// to store answer
string ans = "";
if (d != 10) {
ans += d.ToString();
for (int i = 1; i < n; i++)
ans += '0';
}
else {
if (n == 1)
ans += "Impossible";
else {
ans += '1';
for (int i = 1; i < n; i++)
ans += '0';
}
}
return ans;
}
// Driver code
public static void Main ()
{
int n = 12, d = 3;
Console.WriteLine(findNumber(n, d));
}
}
// This code is contributed by Subhadeep
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to Find N digits
// number which is divisible by D
// Function to return N digits
// number which is divisible by D
function findNumber($n, $d)
{
// to store answer
$ans = "";
if ($d != 10)
{
$ans .= strval($d);
for($i = 1; $i < $n; $i++)
$ans .= '0';
}
else
{
if (n == 1)
$ans .= "Impossible";
else
$ans .= '1';
for($i = 1; $i < $n; $i++)
$ans .= '0';
}
return $ans;
}
// Driver code
$n = 12;
$d = 3;
print(findNumber($n, $d));
// This code is contributed by mits
java 描述语言
<script>
// Javascript program to Find N digits
// number which is divisible by D
// Function to return N digits
// number which is divisible by D
function findNumber(n,d)
{
// to store answer
let ans = "";
if (d != 10) {
ans += (d).toString();
for (let i = 1; i < n; i++)
ans += '0';
}
else {
if (n == 1)
ans += "Impossible";
else {
ans += '1';
for (let i = 1; i < n; i++)
ans += '0';
}
}
return ans;
}
// Driver code
let n = 12, d = 3;
document.write(findNumber(n, d));
// This code is contributed by avanitrachhadiya2155
</script>
Output:
300000000000
版权属于：月萌API www.moonapi.com，转载请注明出处
