pat 1023 Have Fun with Numbers（20 分）

1023 Have Fun with Numbers（20 分）

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

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <map>
 6 #include <stack>
 7 #include <vector>
 8 #include <queue>
 9 #include <set>
10 #define LL long long
11 using namespace std;
12 const int MAX = 30;
13 
14 long long A[15] = {0}, B[15] = {0}, len;
15 char s[MAX], s2[MAX];
16 bool is_equal()
17 {
18     for (int i = 0; i <= 9; ++ i)
19         if (A[i] != B[i]) return false;
20     return true;
21 }
22 
23 void calcS2()
24 {
25     int b = 0, temp[30];
26     for (int i = 0, j = len - 1; i < len; ++ i, -- j)
27     {
28         if (j != -1) b += 2 * (s[j] - '0');
29         temp[i] = b % 10;
30         b /= 10;
31         if (b > 0 && i == len - 1) ++ len;
32     }
33     for (int i = 0, j = len - 1; i < len; ++ i, -- j)
34         s2[j] = char('0' + temp[i]);
35 }
36 
37 int main()
38 {
39 //    freopen("Date1.txt", "r", stdin);
40     scanf("%s", &s);
41     len = strlen(s);
42     for (int i = 0; i < len; ++ i)
43         A[s[i] - '0'] ++;
44     calcS2();
45     len = strlen(s2);
46     for (int i = 0; i < len; ++ i)
47         B[s2[i] - '0'] ++;
48     if (is_equal()) cout <<"Yes" <<endl <<s2 <<endl;
49     else cout <<"No" <<endl <<s2 <<endl;
50     return 0;
51 }