PAT_A1023#Have Fun with Numbers

Source:

PAT A1023 Have Fun with Numbers (20 分)

Description:

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

Keys:

• 大整数运算

Code:

 /*
 Data: 2019-07-11 20:38:07
 Problem: PAT_A1023#Have Fun with Numbers
 AC: 16:20

 题目大意：
 给定K位数，翻倍后，是否为原数的重排列
 */
 #include<cstdio>
 #include<string>
 #include<algorithm>
 #include<iostream>
 using namespace std;

 string Double(string s)
 {
     string ans;
     int carry=,len=s.size();
     for(int i=; i<len; i++)
     {
         int temp = (s[len-i-]-'')*+carry;
         ans.insert(ans.end(), temp%+'');
         carry = temp/;
     }
     while(carry != )
     {
         ans.insert(ans.end(), carry%+'');
         carry /= ;
     }
     reverse(ans.begin(),ans.end());
     return ans;
 }

 int main()
 {
     string s;
     cin >> s;
     string t = Double(s);
     string p = t;
     sort(s.begin(),s.end());
     sort(t.begin(),t.end());
     if(s == t)  printf("Yes\n");
     else        printf("No\n");
     cout << p;

     return ;
 }