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
大数运算, 用字符实现加减
1 #include <iostream> 2 #include <algorithm> 3 #include <string> 4 using namespace std; 5 //会溢出,不能使用简单的加减 6 int main() 7 { 8 string a, b = "", res; 9 cin >> a; 10 int k = 0; 11 for (int i = a.length() - 1; i >= 0; --i) 12 { 13 k = k + (a[i] - '0') + (a[i] - '0'); 14 b += k % 10 + '0'; 15 k /= 10; 16 } 17 if (k > 0) 18 b += k + '0'; 19 res.assign(b.rbegin(), b.rend()); 20 sort(a.begin(), a.end()); 21 sort(b.begin(), b.end()); 22 if (a == b) 23 cout << "Yes" << endl; 24 else 25 cout << "No" << endl; 26 for (auto c : res) 27 cout << c; 28 cout << endl; 29 return 0; 30 }