1023. Have Fun with Numbers (20)

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 differentpermutation. 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:

1234567899

Sample 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];}}}

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