ZOJ--1078：Palindrom Numbers

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思路：

这个题呢首先知道怎么转换进制，就是一直取余。然后判断这个数是否为回文数即可

比较简单不做过多解释

java：

import java.util.Scanner;public class Palindrom_Numbers_1078 {static int arr[]=new int[1000];static int b[]=new int[15];public static void main(String[] args) {Scanner s=new Scanner(System.in);int n;while((n=s.nextInt())!=0){int count=0;for(int i=2;i<=16;i++){int x=n,t=0;while(x!=0){arr[t++]=x%i;x/=i;}            if(judge(t))            b[count++]=i;}if(count>0){System.out.print("Number "+n+" is palindrom in basis");for(int i=0;i<count;i++)System.out.print(" "+b[i]);System.out.println();}elseSystem.out.println("Number "+n+" is not a palindrom");}}public static boolean judge(int k){for(int i=0;i<k/2;i++){if(arr[i]!=arr[k-1-i])return false;}return true;}}

c++：

#include<iostream>using namespace std;int arr[1000];int b[15];bool judge(int k){for(int i=0;i<k/2;i++){if(arr[i]!=arr[k-1-i])return false;}return true;}int main(){int n;while(cin>>n&&n){int co=0;for(int i=2;i<=16;i++){int x=n;int t=0;while(x){arr[t++]=x%i;x/=i;}if(judge(t))    b[co++]=i;}if(co>0){cout<<"Number "<<n<<" is palindrom in basis";for(int i=0;i<co;i++){cout<<" "<<b[i];}cout<<endl;}else    cout<<"Number "<<n<<" is not a palindrom"<<endl;}return 0;}

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