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:1234567899Sample Output:
Yes 2469135798
#include <iostream> #include <string> using namespace std; int main() { string s; int head; cin>>s; int *arr = new int[s.length()]; int *doubleArr = new int[s.length()]; for(int i=0;i<s.length();i++){ arr[i] = s[i]-'0'; } head = arr[0]; int flag = 0; for(int i=s.length()-1;i>=0;i--){ if(flag==0){ if(arr[i]*2>=10){ flag = 1; doubleArr[i] = (arr[i]*2)%10; }else{ doubleArr[i] = (arr[i]*2)%10; flag = 0; } }else{ if((arr[i]*2+1)>=10){ flag = 1; doubleArr[i] = (arr[i]*2+1)%10; }else{ doubleArr[i] = (arr[i]*2+1)%10; flag = 0; } } } int isFunNum = 0; for(int i=0;i<s.length();i++){ int findL = 0; for(int j=0;j<s.length();j++){ if(doubleArr[i]==arr[j]){ findL = 1; arr[j] = -1; break; } } if(findL==0){ isFunNum = 0; break; } isFunNum = 1; } if(isFunNum == 0){ cout<<"No"<<endl; if(doubleArr[0]<=head) cout<<1; for(int i=0;i<s.length();i++){ cout<<doubleArr[i]; } }else{ cout<<"Yes"<<endl; if(doubleArr[0]<=head) cout<<1; for(int i=0;i<s.length();i++){ cout<<doubleArr[i]; } } }