PAT PAT 1023. Have Fun with Numbers (20)

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：给你一个k位的整数，然后将它乘以2，得到结果sum，如果结果sum的组成数字出现的次数和给出的数相同，那么我们就输出Yes，反之，输出No，

然后再把sum输出。
#include 
#include 
#include 
using namespace std;
int main()
{
    char s[30],c[30],d[30];
    int len,i,j,cnt,cnt1[15],cnt2[15],x,flag;
    while(cin>>s)
    {
        len=strlen(s);
        cnt=0;
        for(i=len-1;i>=0;i--)
        {
            x=(s[i]-'0')*2;
            c[i]=x%10+cnt+'0';//把进位后的数组存起来
            cnt=x/10;//进位
        }
        if(cnt==1)//如果在最后一次运算都有进位，那么肯定是错的，因为相同的数字出现了多次
        {
            cout<<"No"<