# 题目

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:
1234567899
Sample Output:
Yes
2469135798

# 源代码

#include <iostream>
#include <string>
#include <vector>

using namespace std;

int countOld[11], countNew[11];

int main(){
string s1;
cin >> s1;
for(int i = 0; i < s1.size(); ++i){
++countOld[(int)(s1[i] - '0')];
}
// * 2
vector<int> v;
const int NUM = 2;
int carry = 0;
for(int i = s1.size() - 1; i >= 0; --i){
int tmp = (s1[i] - '0') * NUM + carry;
carry = tmp / 10;
v.push_back(tmp % 10);
}
if(carry){ //最高位进位情况
v.push_back(carry);
}
// count
for(int i = 0; i < v.size(); ++i){
++countNew[v[i]];
}
bool ifEquals = true;
for(int i = 0; i < 11; ++i){
if(countOld[i] != countNew[i]){
ifEquals = false;
break;
}
}
ifEquals ? printf("Yes\n") : printf("No\n");
for(int i = v.size() - 1; i >= 0; --i){
printf("%d", v[i]);
}
return 0;
}