完整C代码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

以下是完整的C代码实现：

#include <stdio.h>
#include <string.h>

int main() {
    char num[21];
    scanf("%s", num);
    int len = strlen(num);
    char double_num[22];
    int i, j;
    for (i = len - 1, j = 0; i >= 0; i--, j++) {
        int digit = num[i] - '0';
        int double_digit = digit * 2;
        if (double_digit >= 10) {
            double_num[j] = double_digit % 10 + '0';
            j++;
            double_num[j] = '1';
        } else {
            double_num[j] = double_digit + '0';
        }
    }
    double_num[j] = '\0';
    if (strlen(double_num) != len) {
        printf("No\n%s\n", double_num);
        return 0;
    }
    int num_count[10] = {0};
    int double_count[10] = {0};
    for (i = 0; i < len; i++) {
        num_count[num[i] - '0']++;
        double_count[double_num[i] - '0']++;
    }
    for (i = 0; i < 10; i++) {
        if (num_count[i] != double_count[i]) {
            printf("No\n%s\n", double_num);
            return 0;
        }
    }
    printf("Yes\n%s\n", double_num);
    return 0;
}

这里我们同样先读入输入的字符串，然后按位从低到高计算出其对应的双倍数。由于双倍数可能会进位，因此我们需要在输出的字符串中分别填入高位和低位的数字。最后，我们比较原始数和双倍数中每个数字出现的次数，如果不一致，则说明双倍数不是原始数的排列。

需要注意的是，我们在输出字符串时需要判断其长度是否和原始数相等，如果不相等，则说明双倍数的位数比原始数多，此时直接输出即可。