PAT (Advanced Level) 1019 General Palindromic Number (20 分) 【Python/C++】【进制转换】_1019 general palindromic number python-优快云博客

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博客提及PAT（Advanced Level）1019 General Palindromic Number题目，分值20分，并给出了Python和C++两种语言的相关内容，聚焦信息技术领域的编程解题。

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PAT (Advanced Level) 1019 General Palindromic Number (20 分)

Python：

import sys
from math import *
str_in = input()
a = [int(n) for n in str_in.split()]
n = int(a[0])
d = int(a[1])

vis = [0] * 50
k = int(0)
while int(n) > 0:
    k = k + 1
    vis[k] = int(n % d)
    n //= d

flag = 1
for i in range(1, int(k/2 + 1)):
    if vis[i] != vis[k + 1 - i]:
        flag = 0
        break

if flag:
    print("Yes")
    for i in range(1, k):
        print(vis[i],end = ' ')
    print(vis[k])
  
else:
    print("No")
    for i in range(k, 1, -1):
        print(vis[i],end = ' ')
    print(vis[1])

C++：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cmath>
#include<map>
#include<queue>
#include<vector>
using namespace std;
#define ll long long int
#define LL ll
#define INF 0x3f3f3f3f
const int maxn = 1e5 + 10;

int a[maxn];
int n;
int d;
int main() {
	scanf("%d %d", &n, &d);
	int k = 0;
	while (n) {
		a[++k] = n % d;
		n /= d;
	}
	int flag = 1;
	for (int i = 1; i <= k / 2; i++) {
		if (a[i] != a[k - i + 1]) {
			flag = 0; break;
		}
	}
	if (flag) {
		puts("Yes");
		for (int i = 1; i <= k; i++) {
			if (i ^ 1) {
				printf(" ");
			}
			printf("%d", a[i]);
		}
		printf("\n");
	}
	else {
		puts("No");
		for (int i = k; i >= 1; i--) {
			if (i ^ k) {
				printf(" ");
			}
			printf("%d", a[i]);
		}
		printf("\n");
	}
	
	return 0;
}
//_CRT_SECURE_NO_WARNINGS