PTA (Advanced Level) 1019 General Palindromic Number
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 (. Here, as usual, 0for 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 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 is the decimal number and 2 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
解题思路:
本题有多组测试数据,每组数据给出一个十进制数字n,与目标进制b,要求将b转化为b进制后,判断转化完成的数字是否为回文数字。
我们可以写一个函数,将十进制数字n转化为b进制并将其每一位都记录在一个容器num中,之后从num第一位开始到中心为止,判断该位与其对称位置是否全相等即可。若全相等,则目标数字是回文数字,否则不是回文数字。
AC代码
#include <bits/stdc++.h>
using namespace std;
vector<int> num; //用来储存进制转换后数字的容器
void changeRadix(int n, int radix){ //进制转换函数
while(n){
int temp;
temp = (n % radix);
num.push_back(temp);
n /= radix;
}
}
bool judge(){ //回文判断函数
int len = num.size();
for(int i = ; i < len / ; i++){
if(num[i] != num[len - i - ]) //判断该位置与其对称位置是否相等
return false;
}
return true;
}
int main()
{
int n, b;
while(scanf("%d%d", &n, &b) != EOF){ //输入数字n与目标进制b
if(n == ){ //0的所有进制都是回文数
printf("Yes\n");
printf("0\n");
continue;
}
num.clear(); //清空容器num
changeRadix(n, b); //将n转化为b进制
if(judge()){ //判断转换进制后是否为回文数字
printf("Yes\n"); //是则输出Yes
//由于我们是逆序储存转换后的数字的,所以在输出前要将其反转
reverse(num.begin(), num.end());
bool flag = false;
for(auto i : num){ //输出转换后数字
if(flag)
printf(" ");
else
flag = true;
cout << i;
}
printf("\n");
}else{ //不是回文数字输出No,其余操作同上
printf("No\n");
reverse(num.begin(), num.end());
bool flag = false;
for(auto i : num){
if(flag)
printf(" ");
else
flag = true;
cout << i;
}
printf("\n");
}
}
return ;
}
PTA (Advanced Level) 1019 General Palindromic Number的更多相关文章
- PAT (Advanced Level) 1019. General Palindromic Number (20)
简单题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> ...
- PAT (Advanced Level) Practice 1019 General Palindromic Number (20 分) 凌宸1642
PAT (Advanced Level) Practice 1019 General Palindromic Number (20 分) 凌宸1642 题目描述: A number that will ...
- PAT 1019 General Palindromic Number
1019 General Palindromic Number (20 分) A number that will be the same when it is written forwards ...
- PAT 甲级 1019 General Palindromic Number(20)(测试点分析)
1019 General Palindromic Number(20 分) A number that will be the same when it is written forwards or ...
- PAT 1019 General Palindromic Number[简单]
1019 General Palindromic Number (20)(20 分) A number that will be the same when it is written forward ...
- 1019 General Palindromic Number (20 分)
1019 General Palindromic Number (20 分) A number that will be the same when it is written forwards or ...
- PAT 甲级 1019 General Palindromic Number(简单题)
1019. General Palindromic Number (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN ...
- PAT 甲级 1019 General Palindromic Number (进制转换,vector运用,一开始2个测试点没过)
1019 General Palindromic Number (20 分) A number that will be the same when it is written forwards ...
- PAT甲级——1019 General Palindromic Number
A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...
随机推荐
- eclipse-->run as --> maven test 中文乱码
其有一个配置参数forkMode,默认为once,即表示每次运行test时,新建一个JVM进程运行所有test. 这可能会导致乱码问题.首先将forkMode设置为never,即不新建.再运行mvn ...
- WPF 按名称查找控件
1FrameworkElement类FindName方法 使用过程 1.容器控件.RegisterName("Name",要注册的控件) //注册控件 2.容器控件.FindN ...
- djang系列5.5-- 图书管理系统实例
一.表格设计 E-R图 分析图 models.py from django.db import models # Create your models here. class Author(model ...
- SpringMVC中重定向传参数的方法
在spring的一个controller中要把参数传到页面,只要配置视图解析器,把参数添加到Model中,在页面用el表达式就可以取到.但是,这样使用的是forward方式,浏览器的地址栏是不变的,如 ...
- 基于CH340的一键下载电路
一.CH340简介 CH340 是一个 USB 总线的转接芯片,实现 USB 转串口或者 USB 转打印口.CH340是国产芯片,应用场合居多,市场占有率很高.常用的USB转串口芯片还有CP2102. ...
- urllib的使用
1.urllib 中的urlopen urllib.urlopen(url,data) 如果请求是json格式,则data是json.dumps(data_dict)形成的数据,注意,不能在进行url ...
- Python库大全
网络 urllib -网络库(stdlib). requests -网络库. grab – 网络库(基于pycurl). pycurl – 网络库(绑定libcurl). urllib3 – Pyth ...
- Fleury算法
关于为什么不选桥 因为选桥之后会变成两个联通分支,这时由于可能产生的新联通分支不是孤立顶点,他俩都不联通了,那么也就绝对不可能“一笔画”走下来了 关于为什么可以选除桥之外的任意一条边走 本质原因是因为 ...
- python实现计算器功能
import re def strip_operate(exp): # 合并多余的操作符 exp = exp.replace("+-", "-") exp = ...
- swagger注释API详细说明
API详细说明 注释汇总 @RequestMapping此注解的推荐配置 value method produces 示例: @ApiOperation("信息软删除") @Api ...