(PAT 1019) General Palindromic Number (进制转换)

青旅半醒 2022-03-17 01:36 159阅读 0赞

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 ∑i=0k(aibi). 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 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 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:

题目要求判断10进制数转为B进制数后得到的序列是否是回文串

首先是进制转换,使用取余法:

void decimalToRadix(long long num, long long nbase) {
    	while (num != 0) {
    		numlist[index++] = num % nbase;
    		num /= nbase;
    	}
    }

得到序列后,利用双指针判断即可

#define _CRT_SECURE_NO_WARNINGS
    #include <iostream>
    #include <algorithm>
    #include <math.h>
    #include <string>
    const int MAXN = 100010;
    using namespace std;
    long long numlist[MAXN];
    long long index = 0;
    void decimalToRadix(long long num, long long nbase) {
    	while (num != 0) {
    		numlist[index++] = num % nbase;
    		num /= nbase;
    	}
    }
    int main() {
    	long long dnum, dbase;
    	scanf("%lld %lld", &dnum, &dbase);
    	decimalToRadix(dnum, dbase);
    	bool flag = true;
    	long long point1 = 0, point2 = index - 1;
    	while (point1 < point2) {
    		if (numlist[point1] != numlist[point2]) {
    			flag = false;
    			break;
    		}
    		point1++, point2--;
    	}
    
    	if (flag) {
    		cout << "Yes" << endl;
    	}
    	else {
    		cout << "No" << endl;
    	}
    	for (int i = index-1; i >= 0; --i) {
    		printf("%lld", numlist[i]);
    		if (i > 0) cout << " ";
    	}
    
    	system("PAUSE");
    	return 0;
    }

(PAT 1019) General Palindromic Number (进制转换)

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