1019. General Palindromic Number (20)

最新推荐文章于 2021-02-18 10:58:51 发布

转载最新推荐文章于 2021-02-18 10:58:51 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/grglym/p/7638225.html

本文介绍了一个算法，该算法将一个十进制数转换为指定进制的数，并判断转换后的数是否为回文数。文章提供了一段C++代码实现，通过输入一个非负整数N和一个进制b，程序会输出该数在b进制下是否为回文数及其具体的b进制形式。

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A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number N > 0 in base b >= 2, where it is written in standard notation with k+1 digits a_i as the sum of (a_ibⁱ) for i from 0 to k. Here, as usual, 0 <= a_i < b for all i and a_k is non-zero. Then N is palindromic if and only if a_i = a_k-i for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any non-negative decimal integer N and a base b, you are supposed to tell if N is a palindromic number in base b.

Input Specification:

Each input file contains one test case. Each case consists of two non-negative numbers N and b, where 0 <= N <= 10⁹ is the decimal number and 2 <= b <= 10⁹ is the base. The numbers are separated by a space.

Output Specification:

For each test case, first print in one line "Yes" if N is a palindromic number in base b, or "No" if not. Then in the next line, print N as the number in base b in the form "a_k a_k-1 ... a₀". Notice that there must be no extra space at the end of output.

Sample Input 1:

27 2

Sample Output 1:

Yes
1 1 0 1 1

Sample Input 2:

121 5

Sample Output 2:

No
4 4 1

题目大意就是求将一个整数n（0<=n<=10^9）转化成b进制数num，在求num是否是回文数。

#include<iostream>
#include<stdio.h>
using namespace std;
int main(){
	int n,b;
	int num[100];
	int cnt=0;
	int i,j;
	scanf("%d%d",&n,&b);
	if(n == 0){
		cnt = 1;
		num[0]=0;
	}
	for(;n>0;){
		num[cnt]=n%b;
		n=n/b;
		cnt++;
	}
	int flag = 1;
	for(i=0,j=cnt-1;i<j;i++,j--){
		if(num[i]!=num[j]){
			flag = 0;
			break;
		}
	}
	if(flag ==0){
		printf("No\n");
	} 
	else {
		printf("Yes\n");
	}
	for(i=cnt-1;i>=0;i--){
		if(i == 0){
			printf("%d\n",num[i]);
		}
		else {
			printf("%d ",num[i]);
		}
	}
}

转载于:https://www.cnblogs.com/grglym/p/7638225.html