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: |
| 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("\n");
}
return 0;
}
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