PTA 自测-4 Have Fun with Numbers

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探讨一种特殊性质的数字，当该数字乘以2后，其结果中的数字排列仅是原数字中数字的另一种排列。博客详细解析了如何通过C++的高精度算法检查任意不超过20位的正整数是否具有这种特性。

PTA 自测-4 Have Fun with Numbers

题目描述

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

输入格式

Each input contains one test case. Each case contains one positive integer with no more than 20 digits.

输出格式

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

输入输出样例

输入样例#1

1234567899

输出样例#1

Yes
2469135798

题目思路

理解题意，题目意思为原数x2之后，每个数字出现的次数相同为Yes，并且无论Yes/No都要输出x2的结果。
最大长度为20位，所以采用C++的高精度算法来进行计算。

#include<iostream>
#include<string>
#include<vector>
int a[10], b[10];
using namespace std;
int main()
{
	string str;
	cin >> str;
	vector<int> c;
	for (int i = str.size() - 1; i >= 0; i--)c.push_back(str[i] - '0');
	int t = 0;
	for (int i = 0; i < c.size(); i++) {
		c[i] *= 2;
		c[i] += t;
		t = c[i] > 9 ? 1 : 0;
		if (t)c[i] %= 10;
	}
	if (t)c.push_back(1);
	if (c.size() != str.size()) {
		cout << "No" << endl;
		for (int i = c.size() - 1; i >= 0; i--)
			cout << c[i];
		return 0;
	}
	for (int i = 0; i < str.size(); i++) {
		a[str[i] - '0']++;
		b[c[i]]++;
	}
	for (int i = 0; i < 10; i++)
		if (a[i] != b[i]) {
			cout << "No" << endl;
			for (int i = c.size() - 1; i >= 0; i--)
				cout << c[i];
			return 0;
		}
	cout << "Yes" << endl;
	for (int i = c.size() - 1; i >= 0; i--)
		cout << c[i];
	return 0;
}