Despite the glorious fall colors in the midwest, there is a great deal of time to spend while on a train from St. Louis to Chicago. On a recent trip, we passed some time with the following game.

We start with a positive integer S. So long as it has more than one digit, we compute the product of its digits and repeat. For example, if starting with 95, we compute 9 × 5 = 45 . Since 45 has more than one digit, we compute 4 × 5 = 20 . Continuing with 20, we compute 2 × 0 = 0 . Having reached 0, which is a single-digit number, the game is over.

As a second example, if we begin with 396, we get the following computations:

3 × 9 × 6 = 162

1 × 6 × 2 = 12

1 × 2 = 2

and we stop the game having reached 2.

95

396

28

4

40

0

95 45 20 0396 162 12 228 16 6440 0

题意：给出一个数字，将每一位相乘得到下一个数字，知道数字位数为1则停止，输出所有情况

水题，不解释

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;

int main()

{

    int n,t,r,s;

    while(~scanf("%d",&n),n)

    {

        int cnt = 0;

        printf("%d",n);

        if(n>=10)

        {

            while(n)

            {

                t = n;

                s = 1;

                while(t)

                {

                    r = t%10;

                    s*=r;

                    t/=10;

                }

                n = s;

                if(n/10==0)

                {

                    printf(" %d",n);

                    break;

                }

                printf(" %d",s);

            }

        }

        printf("\n");

    }

    return 0;

}