PAT甲级-1023-Have Fun with Numbers(大整数运算+乘2运算)

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

思路：

最高位不出现进位，并且每个数字出现的次数和原有数字出现次数相同，则打印Yes，否则打印No。十分注意，最高位出现进位时要在最高位前面加上一个数字1！！！

代码如下

#include<iostream>
#include<cstring>
using namespace std;
int digit[10];//存储各个原有数字出现的次数 
int main()
{
	char num[25];int up = 0;//up表示进位 
	scanf("%s", num);
	for(int i = strlen(num)-1; i >= 0; i--){
		int tmp = num[i]-'0';
		digit[tmp]++;
		tmp = tmp*2 + up;
		up = 0;//进位置0
		if(tmp >= 10){
			tmp -= 10;
			up = 1;//进位置1
		} 
		num[i] = tmp + '0';
		digit[tmp]--; 
	}
	int flag = 0;
	for(int i = 0; i < 10; i++){
		if(digit[i] != 0) flag = 1;
	}
	printf("%s", (up==1 || flag==1)?"No\n":"Yes\n");
	if(up == 1) printf("1");
	printf("%s", num);
	return 0;
}