PAT甲级1019 回文数

题目

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 the sum of (aibi) for i from 0 to k. 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 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 “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

题解

#include<iostream>
using namespace std;int a[20]={0};int num=0;
void reverse(int n,int d){while(n!=0){a[num++]=n%d;n/=d;}
}int main(){int n,d;scanf("%d %d",&n,&d);reverse(n,d);int flag=0;for(int i=0;i<num/2;i++){if(a[i]!=a[num-1-i]){printf("No\n");flag=1;break;}}if(!flag)  printf("Yes\n");for(int i=num-1;i>=0;i--){printf("%d",a[i]);if(i!=0) printf(" ");}if(num==0) printf("0");return 0;
}

学习到的点是：特殊点 0的问题。有特殊点为0时是Yes，则判断时先考虑输出No的情况。判断时可使用标识flag，只要有错flag设为1，输出No，且终止所有循环判断。标识未被改变，在循环外输出Yes。

还有输出空格时的讲究，可以用i的值控制。

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PAT甲级1019 回文数

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