PAT 1023. Have Fun with Numbers (20)

最新推荐文章于 2024-05-20 09:00:59 发布

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本文介绍了一种简单的方法来检查一个给定的数字在翻倍之后是否仅由其原始数字的不同排列组成。通过列竖式计算的方式进行验证，并提供了一个具体的实现示例。

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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

思路：这题比较简单，列竖式计算即可

#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
int main()
{
  string num;
  vector<int> v;
  vector<int> v1;
  stack<int> s;
  cin>>num;
  for(int i = 0; i < num.size(); i++)
  {
    v.push_back(num[i] - '0');
  }
  int mod = 0;
  for(int i = v.size() - 1; i >= 0; i--)
  {
    s.push((v[i] * 2 + mod) % 10);
    mod = (v[i] * 2 + mod) / 10;
  }
  if(mod != 0)
   s.push(mod);
  string res;
  while(!s.empty())
  {
    res = res + static_cast<char>(s.top() + '0');
    v1.push_back(s.top());
    s.pop();
  }
  sort(v.begin(),v.end());
  sort(v1.begin(),v1.end());
  bool flag = true;
  if(v.size() != v1.size())
   flag = false;
  for(int i = 0; i < v.size(); i++)
    if(v[i] != v1[i])
    {
      flag = false;
      break;
    }
  if(flag)
   cout<<"Yes"<<endl;
  else
   cout<<"No"<<endl;
  cout<<res<<endl;
  return 0;
}