PAT甲级真题 1023 Have Fun with Numbers (20分) C++实现（字符串模拟大数乘法）

题目

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.
Input Specification:
Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.
Output Specification:
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.
Sample Input:
1234567899
Sample Output:
Yes
2469135798

思路

long long的最大值约为9.2 * 10 ^ 18，19位10进制数，显然不够题目中20位的要求。

所以需要用字符串存储数字，从低位到高位模拟乘法过程，新数字位 = 旧数字位*2 + 进位。用string存储结果，为了提高效率，可以+=保存每位，最后得到倒序存储的结果。若最高位还有进位，记得在结果中加上进位“1”，并可认定肯定不符合要求。

接下来判断两个字符串是否有相同元素，遍历其中一个中的字符，在另一个中找与其相同的；若找到，则标记为已匹配，本文用的“*”；若没找到， 则可认定不符合要求。

柳神用的方法比较简便：记录原数字中每个数字的个数，新数字中每出现一个某数字，就把某数字个数减1。最后所有数字个数为0即表示二者包含元素相同。

代码

#include <iostream>
using namespace std;

int main(){
   
   
    string s;