HUNNU11352:Digit Solitaire

Problem description

  

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.

Input

   Each line contains a single integer 1 ≤ S ≤ 100000, designating the starting value. The value S will not have any leading zeros. A value of 0 designates the end of the input.

Output

  For each nonzero input value, a single line of output should express the ordered sequence of values that are considered during the game, starting with the original value.

Sample Input

95
396
28
4
40
0

Sample Output

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("
");
    }

    return 0;
}