Carry large numbers string processing &&

Have Fun with Numbers

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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

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Title understood by analyzing the maximum number of bits is 20 bits, and even the largest long long type case will explode, and therefore to adopt be treated with a string . 
On the code:

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

int main(void){
    char num1[21],num2[21];
    scanf("%s",num1);
    char doublenum[21];
    char tmp;
    strcpy(num2,num1);
    int len = strlen (num2);
     // for each digital multiplier after the 2 stored in an array num2 
    int carrybit = 0 ; // store the bits into each operand

     char midch; 

　　　　// string handling key
     for ( int I len = . 1 ; I> = . 1 ; i-- ) { 
　　　　　　// the algorithm according to the procedure given in column artificial calculated by the equation
        midch=num2[i];
        num2 [I] = ((num2 [I] - ' 0 ' ) * 2 + carrybit)% 10 + ' 0 ' ; // first by 2 plus the carry generated by the previous operation
        carrybit = ((midch- ' 0 ' ) * 2 + carrybit) / 10 ; // calculate this time the carry, then careful not num2 [i], because num2 [i] has changed Error: carrybit = ((num2 [ i] - '0') * 2 + carrybit) / 10

　　} 
　　Num2 [ 0 ] = (num2 [ 0 ] - ' 0 ' ) * 2 + carrybit + ' 0 ' ; // the highest level do not have to carry, if the most significant bit is greater than 9, the following will be given treatment 

    IF (num2 [ 0 ]> ' 9 ' ) {// If the multiplier and the number of digits after the original 2 do not match, it is determined no direct
        printf("No\n");
        printf("%d%d",(num2[0]-'0')/10,(num2[0]-'0')%10);
        printf("%s",&num2[1]);
    }else{
        strcpy(doublenum,num2);
        
　　　　//冒泡排序num1 num2
        for(int i=0;i<len-1;i++){
            for(int j=0;j<len-i-1;j++)
             if(num2[j]>num2[j+1]){
                tmp=num2[j];
                num2 [j] = num2 [j + 1 ];
                num2[j+1]=tmp;
            }
        }
        
        for(int i=0;i<len-1;i++){
            for(int j=0;j<len-i-1;j++)
            if(num1[j]>num1[j+1]){
                tmp=num1[j];
                num1 [j] = num1 [j + 1 ];
                num1[j+1]=tmp;
            }
        }

        if(strcmp(num1,num2)==0){
            printf("Yes\n");
            printf("%s",doublenum);
        }else{
            printf("No\n");
            printf("%s",doublenum);
        }
    }

    return 0;
}

Topic Source:

https://pintia.cn/problem-sets/17/problems/263