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  • A1019. General Palindromic Number

    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
    
     1 #include<cstdio>
     2 #include<iostream>
     3 using namespace std;
     4 int main(){
     5     int N, b, m[34], len = 0;
     6     int tag = 1;
     7     scanf("%d%d", &N, &b);
     8     do{
     9         m[len++] = N % b;
    10         N = N / b;
    11     }while(N != 0);
    12     for(int i = 0, j = len - 1; i <= j; i++, j--){
    13         if(m[i] != m[j]){
    14             tag = 0;
    15             break;
    16         }
    17     }
    18     if(tag == 1)
    19         printf("Yes
    ");
    20     else
    21         printf("No
    ");
    22     printf("%d", m[len - 1]);
    23     for(int i = len - 2; i >= 0; i--)
    24         printf(" %d", m[i]);
    25     cin >>N;
    26     return 0;
    27     
    28 }
    View Code

    总结:

    1、进制转换a进制转b进制时,先转为10进制,再转为b进制。将10进制转为b进制时采用除基取余法,将数字N不断除以基,余数作为结果的低位存储,商则循环除基,直到商为0。输出结果时,最后得到的为最高位。

    2、除基取余法最好使用do while循环,因为有可能被除数为0。

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  • 原文地址:https://www.cnblogs.com/zhuqiwei-blog/p/8439087.html
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