用给定的操作生成数字，检查是否是回文
给定一个整数 N ，任务是通过重复该数字来创建一个字符串，使得结果字符串的长度等于原始数字中数字的总和。 例如:如果数字是 61 ，数字之和是 6 + 1 = 7 ，那么重复 61 后生成的字符串将是 7 长度，即 6161616 。 举例:
输入: N = 10101 输出:是 字符串的长度由数字的总和给出，即 1 + 0 + 1 + 0 + 1 = 3。 所以结果串是“101”，是回文。 输入: N = 123 输出:否 字符串的长度将为 1+2 ^ 3 = 6。 所以结果串是“123123”，不是回文。
进场:
- 求 N 的数字之和，重复这个数字，直到字符串的长度等于这个和。
- 现在检查结果字符串是否是回文。
- 如果是回文，则打印是。
- 否则打印否。
以下是上述方法的实现:
C++
// CPP implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function that returns true if str is a palindrome
bool isPalindrome(string str)
{
int len = str.length();
for (int i = 0; i < len / 2; i++)
{
if (str[i] != str[len - 1 - i])
return false;
}
return true;
}
// Function that returns true if the
// generated string is a palindrome
bool createStringAndCheckPalindrome(int N)
{
// sub contains N as a string
ostringstream out;
out << N;
string result = out.str();
string sub = "" + result, res_str = "";
int sum = 0;
// Calculate the sum of the digits
while (N > 0)
{
int digit = N % 10;
sum += digit;
N = N / 10;
}
// Repeat the substring until the length
// of the resultant string < sum
while (res_str.length() < sum)
res_str += sub;
// If length of the resultant string exceeded sum
// then take substring from 0 to sum - 1
if (res_str.length() > sum)
res_str = res_str.substr(0, sum);
// If the generated string is a palindrome
if (isPalindrome(res_str))
return true;
return false;
}
// Driver code
int main()
{
int N = 10101;
if (createStringAndCheckPalindrome(N))
cout << ("Yes");
else
cout << ("No");
}
// This code is contributed by
// Shashank_Sharma
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function that returns true if str is a palindrome
static boolean isPalindrome(String str)
{
int len = str.length();
for (int i = 0; i < len / 2; i++) {
if (str.charAt(i) != str.charAt(len - 1 - i))
return false;
}
return true;
}
// Function that returns true if the
// generated string is a palindrome
static boolean createStringAndCheckPalindrome(int N)
{
// sub contains N as a string
String sub = "" + N, res_str = "";
int sum = 0;
// Calculate the sum of the digits
while (N > 0) {
int digit = N % 10;
sum += digit;
N = N / 10;
}
// Repeat the substring until the length
// of the resultant string < sum
while (res_str.length() < sum)
res_str += sub;
// If length of the resultant string exceeded sum
// then take substring from 0 to sum - 1
if (res_str.length() > sum)
res_str = res_str.substring(0, sum);
// If the generated string is a palindrome
if (isPalindrome(res_str))
return true;
return false;
}
// Driver code
public static void main(String args[])
{
int N = 10101;
if (createStringAndCheckPalindrome(N))
System.out.println("Yes");
else
System.out.println("No");
}
}
Python 3
# Python 3 implementation of the approach
# Function that returns true if
# str is a palindrome
def isPalindrome(s):
l = len(s)
for i in range(l // 2):
if (s[i] != s[l - 1 - i]):
return False
return True
# Function that returns true if the
# generated string is a palindrome
def createStringAndCheckPalindrome(N):
# sub contains N as a string
sub = "" + chr(N)
res_str = ""
sum = 0
# Calculate the sum of the digits
while (N > 0) :
digit = N % 10
sum += digit
N = N // 10
# Repeat the substring until the length
# of the resultant string < sum
while (len(res_str) < sum):
res_str += sub
# If length of the resultant string exceeded
# sum then take substring from 0 to sum - 1
if (len(res_str) > sum):
res_str = res_str[0: sum]
# If the generated string is a palindrome
if (isPalindrome(res_str)):
return True
return False
# Driver code
if __name__ == "__main__":
N = 10101
if (createStringAndCheckPalindrome(N)):
print("Yes")
else:
print("No")
# This code is contributed by ita_c
C
// C# implementation of the approach
using System ;
class GFG {
// Function that returns true if str is a palindrome
static bool isPalindrome(string str)
{
int len = str.Length;
for (int i = 0; i < len / 2; i++) {
if (str[i] != str[len - 1 - i])
return false;
}
return true;
}
// Function that returns true if the
// generated string is a palindrome
static bool createStringAndCheckPalindrome(int N)
{
// sub contains N as a string
string sub = "" + N, res_str = "";
int sum = 0;
// Calculate the sum of the digits
while (N > 0) {
int digit = N % 10;
sum += digit;
N = N / 10;
}
// Repeat the substring until the length
// of the resultant string < sum
while (res_str.Length< sum)
res_str += sub;
// If length of the resultant string exceeded sum
// then take substring from 0 to sum - 1
if (res_str.Length > sum)
res_str = res_str.Substring(0, sum);
// If the generated string is a palindrome
if (isPalindrome(res_str))
return true;
return false;
}
// Driver code
public static void Main()
{
int N = 10101;
if (createStringAndCheckPalindrome(N))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
// This code is contributed by Ryuga
}
java 描述语言
<script>
// JavaScript implementation of the approach
// Function that returns true
// if str is a palindrome
function isPalindrome(str)
{
let len = str.length;
for (let i = 0; i < len / 2; i++) {
if (str[i] != str[len - 1 - i])
return false;
}
return true;
}
// Function that returns true if the
// generated string is a palindrome
function createStringAndCheckPalindrome(N)
{
// sub contains N as a string
let sub = "" + N, res_str = "";
let sum = 0;
// Calculate the sum of the digits
while (N > 0) {
let digit = N % 10;
sum += digit;
N = N / 10;
}
// Repeat the substring until the length
// of the resultant string < sum
while (res_str.length < sum)
res_str += sub;
// If length of the resultant string exceeded sum
// then take substring from 0 to sum - 1
if (res_str.length > sum)
res_str = res_str.substring(0, sum);
// If the generated string is a palindrome
if (isPalindrome(res_str))
return true;
return false;
}
// Driver code
let N = 10101;
if (createStringAndCheckPalindrome(N))
document.write("Yes");
else
document.write("No");
// This code is contributed by avanitrachhadiya2155
</script>
Output:
Yes
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