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 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

注意点：输出正确的判断应该是carry!=0&&flag  
题目大意： 给出一个长度再20以内的数字字符串， 将其 乘以2， 判断结果中的数字个数是不是和原来一样

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main(){
  string s, s1;
  vector<int> v(10, 0);
  cin>>s;
  s1 = s;
  int carry=0, i;
  for(i=s.size()-1; i>=0; i--){
    //记录0-9中哪些数字出现，以出现次数
    v[s1[i]-'0']++;
    //模拟乘以2的操作
    s[i] += (s[i]-'0' + carry);
    if(s[i]>'9'){
      s[i] -= 10;
      carry = 1;
    }else carry = 0;
  }
  bool flag=true;
  if(carry==1) printf("NO
%d%s", carry, s.str());
  else{  
    for(i=0; i<s.size(); i++){
    if(v[s[i]-'0']>0){
      v[s[i]-'0']--;
    }else{
      printf("NO
%s", s.c_str());
      flag = false;
      break;
    }
  }
  }
  if(flag&&carry!=1) cout<<"Yes
"<<s;
  return 0;
}