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 = 30;

long long A[15] = {0}, B[15] = {0}, len;
char s[MAX], s2[MAX];
bool is_equal()
{
    for (int i = 0; i <= 9; ++ i)
        if (A[i] != B[i]) return false;
    return true;
}

void calcS2()
{
    int b = 0, temp[30];
    for (int i = 0, j = len - 1; i < len; ++ i, -- j)
    {
        if (j != -1) b += 2 * (s[j] - '0');
        temp[i] = b % 10;
        b /= 10;
        if (b > 0 && i == len - 1) ++ len;
    }
    for (int i = 0, j = len - 1; i < len; ++ i, -- j)
        s2[j] = char('0' + temp[i]);
}

int main()
{
//    freopen("Date1.txt", "r", stdin);
    scanf("%s", &s);
    len = strlen(s);
    for (int i = 0; i < len; ++ i)
        A[s[i] - '0'] ++;
    calcS2();
    len = strlen(s2);
    for (int i = 0; i < len; ++ i)
        B[s2[i] - '0'] ++;
    if (is_equal()) cout <<"Yes" <<endl <<s2 <<endl;
    else cout <<"No" <<endl <<s2 <<endl;
    return 0;
}