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 b≥2, where it is written in standard notation with k+1 digits ai as ∑i=0k(aibi). 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 and is also palindromic by definition.
Given any positive decimal integer N and a b, you are supposed to tell if N is a palindromic number in b.
Input Specification:
Each input file contains one test case. Each case consists of two positive numbers N and b, where 0<N≤109 is the decimal number and 2≤b≤109 is the . 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 b, or No if not. Then in the next line, print N as the number in 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
Code
#include <iostream>
#include <vector>
#include <algorithm>
#pragma warning(disable:4996)
using namespace std;
int main()
{
int N, ;
scanf(\"%d%d\", &N, & );
vector<int> result;
if (!N) printf(\"Yes\\n0\\n\");
else
{
while (N)
{
result.push_back(N% );
N /= ;
}
bool but = true;
for (int i = 0; i < result.size(); i++)
{
if (result[i] != result[result.size() - i - 1])
{
but = false;
break;
}
}
printf(but ? \"Yes\\n\" : \"No\\n\");
for (int i = result.size() - 1; i >= 0 ; i--)
{
printf(\"%d\", result[i]);
cout << (i == 0 ? \'\\n\' : \' \');
}
}
return 0;
}
思路
注意不要用string,用vector,如果用了string
10 11
这个样例就炸了。
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