Warm up 14 [B] Digit Solitaire

Digit Solitaire

Time Limit: 3000ms, Special Time Limit:7500ms, Memory Limit:65536KB

Total submit users: 55, Accepted users: 55

Problem 12759 : No special judgement

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 0
396 162 12 2
28 16 6
4
40 0

Problem Source

mcpc 2012

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <queue>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 450005
#define ll long long
#define INF 0x7fffffff
char str[100001];
int n, m;
double ans;
int a[maxn];
int fun(int s){
    int i = 0;
    while (s >= 10){
        printf("%d ", s);
        int x = s;
        s = 1;
        while (x){
            s =s* (x % 10);
            x /= 10;
        }
    }
    return s;
}
int main(){
    int cas = 1;
    while (~scanf("%d", &n)&&n){
        printf("%d\n", fun(n));
    }
    return 0;
}