7-49 Have Fun with Numbers（20 分）

7-49 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

---

#include 
#define MAX_BITS 21

int main(void) {
    int bits = 0, i, num[MAX_BITS], doubled_num[MAX_BITS], carry = 0, double_bits;
    int times[10] = { 0,0,0,0,0,0,0,0,0,0 };  //一个数中每个数字出现的次数（下标表示数字，下标对应的数据为该数字出现的次数）
    int after_times[10] = { 0,0,0,0,0,0,0,0,0,0 };  //该数乘2后每个数字出现的次数
    char temp;

    temp = getchar();
    while (temp >= '0'&&temp <= '9') {  //输入原始数
        num[bits] = temp - '0';
        bits++;
        temp = getchar();
    }
    double_bits = bits;
    for (i = bits - 1; i >= 0; i--) {  //模拟乘法
        doubled_num[i] = num[i] * 2 % 10;
        if (carry == 1) {  //需要进位
            doubled_num[i]++;
            carry = 0;
        }
        if (num[i] * 2 >= 10) {
            carry = 1;
        }
        if (i == 0 && num[i] * 2 >= 10) {  //最高位需要进位
            for (int j = bits; j >= 1; j--) {  //所有数字往后挪一位（腾出最高位）
                doubled_num[j] = doubled_num[j - 1];
            }
            doubled_num[0] = 1;  //最高位为1
            double_bits++;  //乘2后的数位数加1
        }
    }
    for (i = 0; i < bits; i++) {  //统计原始数每个数字出现的次数
        times[num[i]]++;
    }
    for (i = 0; i < double_bits; i++) {  //统计乘2后每个数字出现的次数
        after_times[doubled_num[i]]++;
    }
    for (i = 0; i < 10; i++) {  //判断每个数字出现的次数是否一样
        if (times[i] != after_times[i]) {
            break;
        }
    }
    if (i < 10 || bits != double_bits) {  //每个数字出现次数不完全一样 或 乘2后位数与原始数不一样
        printf("No\n");
        int maxbits;

        if (bits != double_bits) {  //判断乘2后的位数
            maxbits = double_bits;
        }
        else {
            maxbits = bits;
        }
        for (i = 0; i < maxbits; i++) {
            printf("%d", doubled_num[i]);
        }
    }
    else {
        printf("Yes\n");
        for (i = 0; i < bits; i++) {
            printf("%d", doubled_num[i]);
        }
    }
    printf("\n");

    return 0;
}