1023. Have Fun with Numbers (20)

题目：

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 file 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

注意：

1、一定要非常小心进位，特别是最高位也需要进位的时候！！！

2、由于这里的输入可能是10^20量级，所以我直接用了字符数组来计算，先将每一位数字乘以2，再判断是否需要进位，并统计0~9每个数字出现的次数是否与输入相同。

代码：

#include
#include
#include
using namespace std;

int main()
{
	//x:input number
	//x2:double number
	char x[21],x2[21],t;
	//xdigit:appearence time if each digit from 0 to 9
	//it is more convenient to compare by using function strcmp than int[]
	char xdigit[11],x2digit[11];
	memset(xdigit, '0',sizeof (xdigit)); xdigit[10]='\0';
	strcpy(x2digit,xdigit);
	int i=0;
	cin>>x;
	while(x[i]!='\0' )
	{
		++xdigit[x[i]- '0'];//count each digit for x
		x2[i]=(x[i]- '0')*2+'0' ;//double each digit
		++i;
	}
	x2[i]= '\0';
	while(--i)
	{
		if(x2[i]-'0' >9)//
		{++x2[i-1];         x2[i]-=10;}
		++x2digit[x2[i]- '0'];//count each digit for x2 except the first digit
	}
	if(x2[0]-'0' >9)
	{ //if the first digit of x2 is more than 9,that means the length of x2 increases by 1 digit
		cout<< "No"<