Description
The multiplicative persistence of a number is defined by Neil Sloane (Neil J.A. Sloane in The Persistence of a Number published in Journal of Recreational Mathematics 6, 1973, pp. 97-98., 1973) as the number of steps to reach a one-digit number when repeatedly multiplying the digits. Example:

679 -> 378 -> 168 -> 48 -> 32 -> 6.

That is, the persistence of 679 is 6. The persistence of a single digit number is 0. At the time of this writing it is known that there are numbers with the persistence of 11. It is not known whether there are numbers with the persistence of 12 but it is known that if they exists then the smallest of them would have more than 3000 digits.
The problem that you are to solve here is: what is the smallest number such that the first step of computing its persistence results in the given number?

Input
For each test case there is a single line of input containing a decimal number with up to 1000 digits. A line containing -1 follows the last test case.

Output
For each test case you are to output one line containing one integer number satisfying the condition stated above or a statement saying that there is no such number in the format shown below.

Sample Input

0
1
4
7
18
49
51
768
-1

Sample Output

10
11
14
17
29
77
There is no such number.
2688

#include <stdio.h>
#include <string.h>

void main()
{
char Numlong[1001],*Numlong1[1001],temp[1002],temp1[1002],temp2[1002],*result[1001];
int len,i=0,counter=0,j,k,s,l,q;
scanf("%s",Numlong);
while(Numlong[0]!='-')
{
len=strlen(Numlong);
Numlong1[i]=(char *)malloc(len+1);
result[i]=(char *)malloc(len+2);
strcpy(Numlong1[i],Numlong);
i++;
counter++;
scanf("%s",Numlong);
}
for(i=0;i<counter;i++)
{

if(strlen(Numlong1[i])==1)
{
temp[1]='1';
temp[0]=Numlong1[i][0];
strcpy(result[i],temp);
continue;
}
else
{
q=0;
strcpy(temp,Numlong1[i]);
s=0;
j=9;
while(j>1)
{
for(k=0;temp[k]!='\0';k++)
{
s=s*10+temp[k]-'0';
temp1[k]='0'+s/j;
s=s%j;
}
temp1[k]='\0';
if(s==0)
{
if(temp1[0]!='0')
strcpy(temp,temp1);
else
{
for(l=1;temp1[l]!='\0';l++)
temp[l-1]=temp1[l];
temp[l-1]='\0';
}
temp2[q++]='0'+j;
temp2[q]='\0';
}
else {j--;s=0;}
if(strlen(temp)==1&&s==0)
{ temp2[q]=temp[0];
temp2[++q]='\0';
break;
}
}
if(j==1)
result[i][0]='-';
else
strcpy(result[i],temp2);
}
}
for(i=0;i<counter;i++)
{
if(result[i][0]=='-')
printf("There is no such number.\n");
else
{
len=strlen(result[i]);
for(j=len-1;j>=0;j--)
printf("%c",result[i][j]);
printf("\n");
}
}
}

Persistent Numbers

我們如果把一個整數中的各個位數相乘會得到另一個整數，如果不斷的重複這個動作最後會得到一個個位數的數。例如：

679 -> 378 -> 168 -> 48 -> 32 -> 6.
我們說679頑固（persistence）的程度是5，因為他花了5個步驟才變成一個個位數。目前已知存在有頑固程度11的數，我們不知道是否有頑固程度12的數，如果有的話，他的長度一定超過3000位數。

在本問題中給你一個數，請你找出是那個數經由上面的一個步驟（也就是把各個位數相乘）所得到的，這樣的數可能不只一個，請你找出最小的那個。例如：給你 18，你可以找出 29, 36, 63, 92, 233, 323, 332等，但是你的答案應該是 29。

Input

每組測試資料一列。含有一個大於等於0的整數，長度最大到 1000 位數。

-1代表輸入結束，請參考Sample Input。

Output

對每組測試資料輸出一列，如上所述。如果找不到這樣的數，請輸出"There is no such number."。

Sample Input Sample Output
0 10
1 11
4 14
7 17
18 29
49 77
51 There is no such number.
768 2688
-1

今天才发现自己的程序数组越界了 在TC2.0可以运行正常,在VC6.0就出现一大片的"烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫烫",
改正的代码如下
#include <stdio.h>
#include <string.h>

void main()
{
char Numlong[1001],Numlong1[100][1002],temp[1001],temp1[1002],temp2[1002],result[100][1002];
int len,i=0,counter=0,j,k,s,l,q;
scanf("%s",Numlong);
while(Numlong[0]!='-')
{
len=strlen(Numlong);
strcpy(Numlong1[counter++],Numlong);
Numlong1[counter-1][len]='\0';
scanf("%s",Numlong);
}
for(i=0;i<counter;i++)
{
if(strlen(Numlong1[i])==1)
{
temp[1]='1';
temp[0]=Numlong1[i][0];
strcpy(result[i],temp);
result[i][2]='\0';
continue;
}
else
{
q=0;
strcpy(temp,Numlong1[i]);
len=strlen(Numlong1[i]);
temp[len]='\0';
s=0;
j=9;
while(j>1)
{
for(k=0;temp[k]!='\0';k++)
{
s=s*10+temp[k]-'0';
temp1[k]='0'+s/j;
s=s%j;
}
temp1[k]='\0';
if(s==0)
{
if(temp1[0]!='0')
{strcpy(temp,temp1);temp[k]='\0';}
else
{
for(l=1;temp1[l]!='\0';l++)
temp[l-1]=temp1[l];
temp[l-1]='\0';
}
temp2[q++]='0'+j;
}
else {j--;s=0;}
if(strlen(temp)==1&&s==0)
{ temp2[q]=temp[0];
temp2[q+1]='\0';
break;
}
}
if(j==1)
result[i][0]='-';
else
{strcpy(result[i],temp2);result[i][q+1]='\0';}
}
}
for(i=0;i<counter;i++)
{
if(result[i][0]=='-')
printf("There is no such number.\n");
else
{
len=strlen(result[i]);
for(j=len-1;j>=0;j--)
printf("%c",result[i][j]);
printf("\n");
}
}
}