PAT 甲级1023 Have Fun with Numbers (20 分)

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

题目大意：

         给定一个不超过20位的整数，要是将其乘以2后仍旧相同（即出现的数一样）则输出yes并输出乘2后的数。否则输出no。

 

思路：

         大整数用string 处理，方便输入。输入后从后往前处理，设置进位step，进行相应的进位计算。最终将字符串重新反转便得到输出结果。

         判定是否相同则将两个string 排序即可。

参考代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
string s1, s2, s3;
int main(){
	cin >> s1;
	int step = 0, temp;
	for(int i = s1.size() - 1; i >= 0; --i){
		temp = step + stoi(s1.substr(i, 1)) * 2;
		step = temp / 10;
		s2 += temp % 10 + '0';
	}
	if(step)	s2.push_back(step + '0');
	s3 = s2;
	reverse(s3.begin(), s3.end());
	sort(s2.begin(), s2.end());
	sort(s1.begin(), s1.end());
	if(s1 == s2){
		printf("Yes\n");
		cout << s3;
	}
	else{
		printf("No\n");
		cout << s3;
	}
	return 0;
}