Have Fun with Numbers(C语言)

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

#include <stdio.h>
#include <math.h>
#define SIZE 20
int main()
{
    int num[SIZE],double_num[SIZE],bit=0; //bit记录原始数位数
    int digit[10]={0,0,0,0,0,0,0,0,0,0}; //digit[i]代表输入数n中数字i出现的次数
    int flag=1; //flag为1时，输出yes,为0时，输出no
    char temp;
    temp = getchar();
    while (temp >= '0'&&temp <= '9') {  //输入原始数
        num[bit] = temp - '0';
        digit[num[bit]]++;  //记录原始数每个数字出现的次数
        bit++;
        temp = getchar();
    }
    for(int i=bit-1;i>=0;i--){     //原始数每位乘以2得到新数组
        double_num[i]=num[i]*2;
    }
    //处理进位，double_num[0]不做进位处理，若>10，则判断flag为0
    for(int i=bit-1;i>0;i--){
        if(double_num[i]/10!=0){
            double_num[i]=double_num[i]%10;
            double_num[i-1]+=1;
        }
        digit[double_num[i]]--;
    }
    //判断原始数double后，是否位数与之前相同，若不同则flag为0
    if(double_num[0]/10!=0){
        flag=0;
    }else{
        digit[double_num[0]]--;
        for(int j=0;j<10;j++){
            if(digit[j]!=0){
                flag=0;
                break;
            }
        }
    }
    //输出结果
    if(flag==0){
        printf("No\n");
    }else{
        printf("Yes\n");
    }
    for(int i=0;i<bit;i++){
        printf("%d",double_num[i]);
    }
    return 0;
}