PAT 甲级 1023 Have Fun with Numbers

1023 Have Fun with Numbers （20 point(s)）

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

Experiential Summing-up

This question seems can not be solved by long long integer because of overflow, so we must using big number to store and calculate it. The other is so easy~That's all. 

(The purpose of using English to portray my solution is that to exercise the ability of my expression of English and accommodate PAT advanced level's style.We can make progress together by reading and comprehending it. Please forgive my basic grammar's and word's error. Of course, I would appreciated it if you can point out my grammar's and word's error in comment section.( •̀∀•́ ) Furthermore, Big Lao please don't laugh at me because I just a English beginner settle for CET6    _(:з」∠)_  )

Accepted Code

#include <cstdio>
#include <map>
#include <algorithm>
#include <cstring>
using namespace std;

struct bign
{
	int len;
	int d[30];
	bign()
	{
		len=0;
		memset(d,0,sizeof(d));
	}
};
bign convert(char s[])
{
	bign a;
	a.len=strlen(s);
	for(int i=a.len-1;i>=0;--i)
	{
		a.d[a.len-1-i]=s[i]-'0';
	}
	return a;
}
bign mul(bign a)
{
	int carry=0;
	bign b;
	for(int i=0;i<a.len;++i)
	{
		int temp=a.d[i]*2+carry;
		b.d[b.len++]=temp%10;
		carry=temp/10;
	}
	while(carry!=0)
	{
		b.d[b.len++]=carry%10;
		carry/=10;
	}
	return b;
}
bool compare(bign a,bign b)
{
	if(a.len!=b.len)
		return false;
	int a1[10]={0},b1[10]={0};
	for(int i=0;i<a.len;++i)
	{
		++a1[a.d[i]];
		++b1[b.d[i]];
	}
	for(int i=0;i<10;++i)
	{
		if(a1[i]!=b1[i])
			return false;
	}
	return true;
}
int main()
{
	char s[30];
	scanf("%s",s);
	bign a=convert(s);
	bign b=mul(a);
	printf("%s\n",compare(a,b)?"Yes":"No");
	for(int i=b.len-1;i>=0;--i)
	{
		printf("%d",b.d[i]);
	}
	return 0;
}