寻找并输出11~999之间的数m,它满足m.m2和m3均为回文数. 回文:各位数字左右对称的整数. 例如:11满足上述条件 112=121,113=1331 判断一个数是否是回文数的方法:求该数的反序数,若反序数和原数相等,则为回文数,否则不是回文数. 例如:121的反序数是121,所以121是回文数 123的反序数是321,所以123不是回文数 C++代码如下: #include<iostream> using namespace std; bool symm(int m) { int i
寻找最近的回文数 给定一个整数 n ,你需要找到与它最近的回文数(不包括自身). "最近的"定义为两个整数差的绝对值最小. 示例 1: 输入: "123" 输出: "121" 注意: n 是由字符串表示的正整数,其长度不超过18. 如果有多个结果,返回最小的那个. public class Solution { public String mirroring(String s) { String x = s.substring(0, (s.len
Given an integer n, find the closest integer (not including itself), which is a palindrome. The 'closest' is defined as absolute difference minimized between two integers. Example 1: Input: "123" Output: "121" Note: The input n is a p
一个b站上的朋友问我,怎么返回五位数的回文数的个数. 我首先百度回文数的概念,就是正读和倒读都一样的数字,例如:10001,99899 等等 数字的位数拆分一头雾水,思来想去,用字符串的方法完美解决! count = 0 for i in range(10000, 100000): a = str(i) if a[0] == a[-1] and a[1] == a[-2]: count +=1 print(i) print(count)
题目链接:http://acm.swust.edu.cn/problem/797/ Time limit(ms): 1000 Memory limit(kb): 10000 Description Palindromes are numbers that read the same forwards as backwards. The number 12321 is a typical palindrome. Given a number base B (2 <= B <= 20 base
1.打擂台 简单的小代码,打擂台.纪念下过去,祝福下新人. public static void main(String[] args){ int[] ld = {1,4,2,10,8,9,5}; int max = ld[0]; for(int i = 1;i<ld.length;i++){ if(ld[i]>max){ max=ld[i]; } } System.out.print(max); } 2.冒泡排序 还是简单的小代码,冒泡排序.纪念下过去,祝福下新人. public stati
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward. Example 1: Input: 121 Output: true Example 2: Input: -121 Output: false Explanation: From left to right, it reads -121. From right to
