PAT-1023-Have Fun with Numbers

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 file 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

【解析】
这道题的意思就是一个数问你这个数的两倍所得到的的数1-9出现的次数和所给的数1-9出现的次数一不一样。

#include
#include
#include
#include
using namespace std;
int main()
{
    int a[10]={0},b[10]={0},i,jinwei,n,flag=0,sum;
    char c[50],d[50];
    cin>>c;
    n=strlen(c);
    for(i=0;i=0;i--)
    {
        sum=(c[i]-'0')*2+jinwei;
        if(sum>=10)//要进位了
        {
            d[i]=sum-10+'0';
            jinwei=1;
        }
        else
        {
            d[i]=sum+'0';
            jinwei=0;
        }
    }
    if(jinwei==1)
    {
        b[1]++;
    }//表示最高位还有一个1
    for(i=0;i