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

回文数判断与基转换

最新推荐文章于 2025-03-15 18:26:50 发布

原创最新推荐文章于 2025-03-15 18:26:50 发布 · 186 阅读

0 ·

CC 4.0 BY-SA版权

PAT-甲级专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个C++程序，该程序能够将一个十进制整数转换为指定的进制表示，并判断该表示是否为回文数。程序首先读取十进制数N和目标进制b，然后进行基转换并检查转换后的数字是否具有回文特性。

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 the sum of (a~i~b^i^) for i from 0 to k. 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 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 <= 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

题意：

输入一个十进制得数，将它转换成K进制，并判断其K进制数是否为回文数，如果是就打印Yes,不是就打印No

C++:

#include<cstdio>
#include<cstring>   // strrev为字符串的倒序 

bool isPalindromic(int a[],int num){  
	for(int i = 0;i <= num/2;i++){
		if(a[i] != a[num-i-1])
			return false;
	}	
	return true;
}

int main(){
	int N,k;
	int a[1000];
	scanf("%d%d",&N,&k);   //十进制 N 转化成 k进制数后判断是否是回文
	
	//修改几进制 
	int t = N;  
	int count=0;  //为位数，接下来打印循环的次数 
	
	do{
		a[count++] = t % k;    // 将正整数的最低位保存到数组a中
        t /= k;            // 将正整数的次低位变为最低位      
	}while(t) ; //这样的话 t为 0 也会被执行一次 
	 
	if(isPalindromic(a,count))
		printf("Yes\n");		
	else
		printf("No\n");
		
	for(int i = count-1;i >= 0;i--){
		printf("%d",a[i]);
		if(i) 	
			printf(" "); 
	}
	return 0;
}