PAT A1019 General Palindromic Number （20 分）——回文，进制转换

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 ∑​i=0​k​​(a​i​​b​i​​). 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 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 "a​k​​ a​k−1​​ ... a​0​​". 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

鸣谢网友“CCPC拿不到牌不改名”修正数据！

 

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
#include <string>
#include <set>
#include <map>
using namespace std;
const int maxn = 10000;
int s[maxn] = { 0 }, s2[maxn] = { 0 };
int main() {
    int n, b;
    cin >> n >> b;
    
    if (n == 0) {
        printf("Yes
0");
        return 0;
    }
    int i = 0;
    while (n != 0) {
        s[i] = n % b;
        s2[i] = s[i];
        n /= b;
        i++;
    }
reverse(s, s+i);
    int flag = 0;
    for (flag; flag < i; flag++) {
        if (s[flag] != s2[flag]) {
            break;
        }
    }
    if (flag == i) printf("Yes
");
    else printf("No
");
    
    int j;
    for (j = 0; j < i-1; j++) {
        printf("%d ", s[j]);
    }
    printf("%d", s[i-1]);
    system("pause");
}

注意点：不能用string来做，因为b可能会很大，用string无法存储太大的数字，最多就255，还是要开一个int数组。一开始觉得用string超容易，只要存进来然后reverse一下直接==比较就ok了，发现有两个测试点一直过不了，难。