pat甲级1019

题目链接

                                   1019 General Palindromic Number （20 分）

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​​

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

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

题目意思:给你两个数a,b问你a在b进制下是不是回文数，是的话输出Yes，不是的话.输出No,并且都输出转换后的结果。

思路:

wa代码:可能会出现f(15)51这种特殊情况，而此时不是回文串，所以这个代码不行。

#include
#include
#include
#include
#define in(x) scanf("%d",&x)
#define out(x) printf("%d",x)
#define per(i,a,b) for(i=a;i=0;i--){
		b+=a[i];
	}
	if(a==b) return true;
	else return false;
}
int main()
{
	string str;
	int a,b;
	in(a),in(b);
	do{
		str+=a%b+'0';
		a/=b;
	}while(a!=0);//用do while而不用while的原因是0也要记录进去 
	if(huiwenjudge(str)) printf("Yes\n");
	else printf("No\n");
	for(int i=str.size()-1;i>=0;i--){
		if(!i) cout<

ac代码:

#include
#include
#include
#define in(x) scanf("%d",&x)
#define out(x) printf("%d",x)
using namespace std;
const int maxn=9e7+5;
int k[maxn];
int main()
{
	int a,b,len=0,i,j;
	in(a),in(b);
	if(!a){
		printf("Yes\n0");
	}
	else{
		while(a){
			k[len++]=a%b;
			a/=b;
		}
		bool flag=true;
		for(i=0,j=len-1;i=0;i--){
			if(i) printf("%d ",k[i]);
			else printf("%d",k[i]);
		}
	}
}