PAT_A1023#Have Fun with Numbers

Source:

PAT A1023 Have Fun with Numbers (20 分)

Description:

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 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

Keys:

• 大整数运算

Code:

/*
Data: 2019-07-11 20:38:07
Problem: PAT_A1023#Have Fun with Numbers
AC: 16:20

题目大意：
给定K位数，翻倍后，是否为原数的重排列
*/
#include<cstdio>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;

string Double(string s)
{
    string ans;
    int carry=0,len=s.size();
    for(int i=0; i<len; i++)
    {
        int temp = (s[len-i-1]-'0')*2+carry;
        ans.insert(ans.end(), temp%10+'0');
        carry = temp/10;
    }
    while(carry != 0)
    {
        ans.insert(ans.end(), carry%10+'0');
        carry /= 10;
    }
    reverse(ans.begin(),ans.end());
    return ans;
}

int main()
{
    string s;
    cin >> s;
    string t = Double(s);
    string p = t;
    sort(s.begin(),s.end());
    sort(t.begin(),t.end());
    if(s == t)  printf("Yes\n");
    else        printf("No\n");
    cout << p;


    return 0;
}