Have Fun with Numbers

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>
int length(long long a);
void bits(int a[],long long b);
void judgement(int a[],int b[],int len);
int main(){
    int j,len;

    long long num;

    int numl[10];

    for(j=0;j<10;j++){
        numl[j] = 0;
    }

    scanf("%lld",&num);

    len = length(num);

    int arr[len];

    bits(arr,num);

    judgement(arr,numl,len);

    num *=2;

    len = length(num);

    int arr_[len];

    bits(arr_,num);

    int rel = 1;

    for(j=0;j<len;j++){
        int n;
        for(n=0;n<10;n++){
            if(numl[n] == 0){
                if(arr_[j] == n){
                    rel = 0;
                }
            }
        }
    }

    if(rel){
        printf("Yes\n");
        printf("%lld",num);
    }else{
        printf("No\n");
        printf("%lld",num);
    }

    return 0;
}
int length(long long a){
    int cnt = 0;
    for( ; ; ){
        a /= 10;
        cnt++;
        if(a == 0)
        break;
    }
    return cnt;
}

void bits(int a[],long long b){
    int i = 0;
    for(i=0; ;i++){
        a[i] = b%10;
        b /= 10;
        if(b == 0)
        break;
    }
}

void judgement(int a[],int b[],int len){
    int i = 0;
    for(i=0;i<len;i++){
        switch(a[i]){
            case 0:
            b[0] = 1;
            break;
            case 1:
            b[1] = 1;
            break;
            case 2:
            b[2] = 1;
            break;
            case 3:
            b[3] = 1;
            break;
            case 4:
            b[4] = 1;
            break;
            case 5:
            b[5] = 1;
            break;
            case 6:
            b[6] = 1;
            break;
            case 7:
            b[7] = 1;
            break;
            case 8:
            b[8] = 1;
            break;
            case 9:
            b[9] = 1;
            break;
        }
    }
}

超过unsigned long，输入中有数字不在结果中，但结果数字都在输入里
最长串，10个数字全出现
最长串全是9
这三种情况显示答案错误，目前自己没有找出来问题

意思是k位数字翻倍后还由1-k组成?
案例里不超过20位数字,很明显longlong存不下,这里用字符串存会比较好

#include<stdio.h>

int a[10];
int b[10];
int main()
{
    long long n;
    scanf("%lld",&n);
    long long ans=n;
    while(n>0)
    {
        a[n%10]++;
        n/=10;
    }
    ans*=2;
    n=ans;
        while(n>0)
    {
        b[n%10]++;
        n/=10;
    }
    int flag=0;
    for(int i=0;i<10;i++)
    {
        if(a[i]!=b[i]) flag=1;
        //cout<<" "<<a[i]<<endl;
    }
    if(flag==0) printf("Yes\n");
    else printf("No\n");
printf("%lld",ans);
    return 0;
    
}