PAT_自测4_Have Fun with Numbers

题目： https://pta.patest.cn/pta/test/17/exam/4/question/263

自测-4 Have Fun with Numbers   (20分)

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 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

题意&解题；  case是 不超过20位，所以自带数据类型不足，需要写一个用数组表示的大数加乘函数。

/**
* https://pta.patest.cn/pta/test/17/exam/4/question/263
* 数据结构练习 00-自测4. Have Fun with Numbers (20)
*/

#include 
#include 
#include 
#include 
 
using namespace std;

string multipleNum(string str, int n);
bool compareStr(string str1, string str2);


int main(int argc, char const *argv[])
{
	
	string str, result;
	cin >> str;
    // str = "012345678990";

	result = multipleNum(str, 2);

    bool isSame = compareStr(str, result);
    if(isSame){
        cout<<"Yes\n"<=0; pos--){
        int num = (int)(str[pos] - '0');           //str[pos]是char格式，用ASCII记录
    	
        int result = num * n + addIn;
   		addIn = result/10;
    	result = result%10;

        strReturn.insert((string::size_type)0,1,result+'0');
    }
    if(0!=addIn){
        strReturn.insert((string::size_type)0,1,addIn+'0');
    }
    return strReturn;
}

bool compareStr(string str1, string str2){
    if(str1.size() != str2.size())
        return false;
    
    int length = str1.size();
    int count1[10] = {0};
    int count2[10] = {0};

    for( int i = 0 ; i < length ; i++ )
    {   //对str计算，最大length为str.size()
        count1[str1[i]-'0']++;
        count2[str2[i]-'0']++;
    }
    for( int i = 0 ; i < 10 ; i++ )
    {   //对array比较，最大length为10 - [0,9]
        // cout<