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 add(int n) {return 1<<n;}void compare(string str) {int len = str.length();int carry = 0;int tmp;int sign1 = 0;int sign2 = 0;while(len) {sign1 += add(str[len-1] – '0');tmp = (str[len-1] – '0') * 2;str[len-1] = '0' + tmp%10 + carry;sign2 += add(str[len-1] – '0');carry = (tmp + carry)/10;len–;}if(sign1 == sign2) {cout<<"Yes"<<endl;} else {cout<<"No"<<endl;}if(carry == 0) {cout<<str<<endl;} else {cout<<carry<<str<<endl;}}int main() {string in;cin>>in;compare(in);return 0;}
，好好的管教你自己，不要管别人。
