1019 General Palindromic Number (20分)甲级真题

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本文介绍了一种算法，用于判断一个正整数在特定进制下是否为回文数，并详细解释了其工作原理和实现步骤。通过将输入的十进制数转换为目标进制，该算法检查转换后的数是否从前往后读和从后往前读相同。

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 ai as ∑i=0k(aibi). Here, as usual, 0≤ai<b for all i and ak is non-zero. Then N is palindromic if and only if ai=ak−i for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any positive 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 positive 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 "ak ak−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：判断输入N的d进制数是否是回文数，是就输出Yes，否则输出No，第二行输出N经过d进制转换的数

#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main() {
    int n, d, flag = 0;
    vector<int> v;
    cin >> n >> d;
    while (n) {
        v.push_back(n % d);
        n /= d;
    }
    for (int i = 0; true; i++) {
        if (i >= v.size() - i - 1) break;
        if (v[i] != v[v.size() - i - 1]) flag = 1;
    }
    if (flag) cout << "No\n";
    else cout << "Yes\n";
    for (int i = 0; i < v.size(); i++) {
        if (i != 0) cout << " ";
        cout << v[v.size() - i - 1];
    }
    return 0;
}