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

 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <vector>
 #include <queue>
 #include <set>
 #define LL long long
 using namespace std;
 const int MAX = ;

 long long A[] = {}, B[] = {}, len;
 char s[MAX], s2[MAX];
 bool is_equal()
 {
     for (int i = ; i <= ; ++ i)
         if (A[i] != B[i]) return false;
     return true;
 }

 void calcS2()
 {
     int b = , temp[];
     for (int i = , j = len - ; i < len; ++ i, -- j)
     {
         if (j != -) b +=  * (s[j] - '');
         temp[i] = b % ;
         b /= ;
         if (b >  && i == len - ) ++ len;
     }
     for (int i = , j = len - ; i < len; ++ i, -- j)
         s2[j] = char('' + temp[i]);
 }

 int main()
 {
 //    freopen("Date1.txt", "r", stdin);
     scanf("%s", &s);
     len = strlen(s);
     for (int i = ; i < len; ++ i)
         A[s[i] - ''] ++;
     calcS2();
     len = strlen(s2);
     for (int i = ; i < len; ++ i)
         B[s2[i] - ''] ++;
     if (is_equal()) cout <<"Yes" <<endl <<s2 <<endl;
     else cout <<"No" <<endl <<s2 <<endl;
     return ;
 }