1023 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 <bits/stdc++.h>
using namespace std;
struct bign
{

    int d[21];
    int len;
    bign()
    {
        memset(d,0,sizeof(d));
        len=0;
    }
};
bign change(char str[])
{
    bign a;
    a.len=strlen(str);
    for(int i=0; i<a.len; i++)
    {
        a.d[i]=str[a.len-i-1]-'0';
    }
    return a;
}
bign multi(bign a,int b)
{
    bign c;
    int carry=0;
    for(int i=0; i<a.len; i++)
    {
        int temp=a.d[i]*b+carry;
        c.d[c.len++]=temp%10;
        carry=temp/10;
    }
    while(carry!=0)
    {
        c.d[c.len++]=carry%10;
        carry/=10;
    }
    return c;
}
bool Judge(bign a,bign b)
{
    if(a.len!=b.len)return false;
    int count[10]= {0};
    for(int i=0; i<a.len; i++)
    {
        count[a.d[i]]++;
        count[b.d[i]]--;
    }
    for(int i=0; i<10; i++)
    {
        if(count[i]!=0)
            return false;
    }
    return true;
}
void print(bign a)
{
    for(int i=a.len-1; i>=0; i--)
    {
        printf("%d",a.d[i]);
    }
}
int main()
{
    char str[21];
    cin.getline(str,21);
    bign a=change(str);
    bign mul=multi(a,2);
    if(Judge(a,mul)==true)printf("Yes\n");
    else printf("No\n");
    print(mul);
    return 0;
}