zoukankan      html  css  js  c++  java
  • 1019. General Palindromic Number (20)

    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 ai as the sum of (aibi) for i from 0 to k. 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 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 <= 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 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
    #include<cstdio>
    int judge(int ans[],int num){
        for(int i = 0; i <= num/2; i++){
            if(ans[i] != ans[num - i -1])
            return 0;
        }
        return 1;
    }
    int main(){
        int n,b,ans[40],num = 0;
        scanf("%d%d",&n,&b);
        do{
            ans[num++] = n % b;
            n /= b;
        }while(n != 0);
         
        if(judge(ans,num)) printf("Yes
    ");
        else printf("No
    ");
        for(int i = num-1; i >= 0; i--){
            printf("%d",ans[i]);
            if(i != 0) printf(" ");
        }    
        return 0;
    }
    #include<cstdio>
    bool judge(int ans[],int num){
        for(int i = 0; i <= num/2; i++){
            if(ans[i] != ans[num - i -1])
            return false;
        }
        return true;
    }
    
    int main(){
        int n,b,ans[40],num = 0;
        scanf("%d%d",&n,&b);
        do{
            ans[num++] = n % b;
            n /= b;
        }while(n != 0);
        bool flag = judge(ans,num);
        if(flag == true) printf("Yes
    ");
        else printf("No
    ");
        for(int i = num-1; i >= 0; i--){
            printf("%d",ans[i]);
            if(i != 0) printf(" ");
        }    
        return 0;
    }
    //18.8.8
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    const int maxn = 10000;
    int num[maxn],len = 0;
    
    bool isPalindromic(int n, int radix){
        do{
            num[len++] = n % radix;
            n /= radix;
        }while(n > 0);
        for(int i = 0; i < len/2; i++){
            if(num[i] != num[len - 1 - i]) return false;
        }
        return true;
    }
    
    int main(){
        int n,b;
        scanf("%d%d",&n,&b);
        if(isPalindromic(n,b)){
            printf("Yes
    ");
        }else{
            printf("No
    ");
        }
        for(int i = len - 1; i >= 0; i--){
                printf("%d",num[i]);
                if(i > 0) printf(" ");
            }
        return 0;
    }
  • 相关阅读:
    js中return的作用
    jstl标签详解总结
    css——overflow属性用法
    oracle数据批处理
    SQL Server 2000/2005 分页SQL — 单条SQL语句
    dataset和实体类 之间的转换
    barmanager设置
    C#集合类(HashTable, Dictionary, ArrayList)与HashTable线程安全
    comboboxEdit 不能输入,只能选择
    C#在父窗口中调用子窗口的过程(无法访问已释放的对象)
  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/8540697.html
Copyright © 2011-2022 走看看