1019 General Palindromic Number （20 分)

2019-03-10

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=0k(aib**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≤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 bin the form “a**k a**k−1 … a0”. Notice that there must be no extra space at the end of output.

Sample Input 1:

27 2

Sample Output 1:

1 2	Yes 1 1 0 1 1

Sample Input 2:

121 5

Sample Output 2:

1
2

No
4 4 1

鸣谢网友“CCPC拿不到牌不改名”修正数据！

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一个十进制数及进制，要求判断该进制下该数是否为回文数

分析

使用一个vector<int>，纪录转换后的各位数字，判断是否回文即可。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <cstring>
#include <cmath>

using namespace std;
vector<int> v1;
void change(int n,int radix){
    if(n==0) {
        v1.push_back(0);
        return;
    }
    int m=0;
    while(n>=pow(radix,m))m++;
    for(int i=m-1;i>=0;i--){
        int base=pow(radix,i);
        int d=n/base;
        n-=d*base;
        v1.push_back(d);
    }
    return ;
}
int main() {
    int n,b,l;
    bool is=true;
    cin>>n>>b;
    change(n,b);
    l=v1.size();

    for(int i=0;i<l/2;i++){
        if(v1[i]!=v1[l-1-i])
            is=false;
    }
    if(is)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
    cout<<v1[0];
    for(int i=1;i<l;i++)
        cout<<" "<<v1[i];
    
}

其他

这里存储必须使用vector，而不能使用string。若采用string，因为转换后的数字是采用十进制表示的（例如16进制中的12，计为12而不是c。因为

1.需要转换为字母很烦

2.第二进制数大了之后字母也不够用。

如果采用vector，则12为一个整体，如采用string，则12变为2个数了，显然达不到判断回文的目的。