自测-4 Have Fun with Numbers

这题不难，主要看了一下，网上好像没有按我这脑回路做的，就记一下吧。

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

/**<  https://pintia.cn/problem-sets/17/problems/263*/
#include 
#include 
#include 

/**< 判断数字出现次数 */
int timesOfNum(int a[], int len, int n);

int main()
{
    bool isFun = true;
    int maxDigit = 21;
    int len = 0;
    char a[maxDigit];
    int b[maxDigit];
    int c[maxDigit];
    int i, carry, tmp;
    carry = tmp = 0;

    /**< 先以字符串形式读入数字 */
    scanf("%s", a);

    /**< 将字符串形式的数字转化为数组b储存 */
    for(i=0; a[i]!='\0'; i++)
    {
        b[i] = a[i] - '0';
        len++;
    }

    /**< 将数组b乘以2，存入新数组c */
    for( i=len-1; i>=0; i-- )
    {
        tmp = b[i]*2+carry;
        c[i] = tmp%10;
        carry = tmp/10;
    }
    c[len] = carry;

    /**< 对比结果 */
    if(carry)
    {
        isFun = false;
    }
    for(i=0; i<=9; i++)
    {
        if(timesOfNum(b, len, i) != timesOfNum(c, len, i))
        {
            isFun = false;
//            printf("checking num:%d\n", i);
//            printf("b:%d\tc:%d\n", timesOfNum(b, len, i), timesOfNum(c, len, i));
        }
    }
    
    /**< 输出部分 */
    if(isFun)
        printf("Yes\n");
    else
        printf("No\n");
    if(carry)
        printf("%d", c[len]);
    for(i=0; i