1019 General Palindromic Number (20分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

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 (ai bi ). 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≤10 ^9 is the decimal number and 2≤b≤10 ^9 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

题意： 给一个十进制数a，进制b。要求将a转化为b进制，判断转化后的数是否为回文数。

while+除法+取余求进制数。存入vector后翻转，比较翻转前后是否相等。

#include<bits/stdc++.h>
using namespace std;
int main() {int a, b; cin >> a >> b;vector<int>v1, v2;        //存储翻转前和翻转后的最终结果while(a) {v1.push_back(a%b);a /= b;}v2 = v1; reverse(v1.begin(), v1.end());cout << (v1==v2 ? "Yes" : "No") << '\n';for(int i = 0; i < v1.size(); i++) {if(i != 0) putchar(' ');cout << v1[i];} return 0;
}

耗时：

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1019 General Palindromic Number (20分)_18行代码AC

立志用最少的代码做最高效的表达

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