1023 Have Fun with Numbers

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

 

题意:

  将一个数字翻倍后,判断该数字是不是原来数字的重新排列。

思路:

  大数加法。

Code:

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int main() {
 6     string s;
 7     cin >> s;
 8     int len = s.length();
 9     int carry = 0;
10     map<int, int> m;
11     for (int i = 0; i < s.length(); ++i) m[s[i] - '0']++;
12     for (int i = len - 1; i >= 0; --i) {
13         s[i] += s[i] + carry - '0';
14         carry = 0;
15         if (s[i] > '9') {
16             s[i] -= 10;
17             carry = 1;
18         }
19     }
20     if (carry == 1) s = '1' + s;
21     int flag = 0;
22     for (int i = 0; i < s.length(); ++i)
23         if (m[s[i] - '0'] > 0)
24             m[s[i] - '0']--;
25         else
26             flag = 1;
27     if (flag == 0)
28         cout << "Yes" << endl;
29     else
30         cout << "No" << endl;
31     cout << s << endl;
32     return 0;
33 }

 

posted @ 2020-06-19 14:29  Veritas_des_Liberty  阅读(229)  评论(0编辑  收藏  举报