/*Alexandra and A*B Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 325 Accepted Submission(s): 70
Problem Description
Alexandra has a little brother. He is new to programming. One day he is solving the following problem: Given two positive integers A and B, output A*B.
This problem is even easier than the last one. Alexandra can't wait to give him another task: Given a positive integer A and a string S(S only contains numbers.), find the minimum positive integer B, such that S is a substring of T, where T is the decimal notation of A*B.
See the sample for better understanding.
Note: S can contain leading zeros, but T can't.
Input
There are multiple test cases (no more than 500). Each case contains a positive integer A and a string S.
A≤10,000,1≤|S|≤8.
Output
For each case, output the required B. It is guaranteed that such B always exists.
To C++ programmers: if you want to output 64-bit integers, please use "%I64d" specifier or cout.
Sample Input
6 8
96 19
2 0086
1 1
Sample Output
3
2
5043
1
Source
BestCoder Round #19 */
#include<stdio.h>
#include<string.h>
int main()
{
long long a;
while((scanf("%d",&a))!=EOF)
{
long long o=0,b=1,n,f,p,t,i,j;char s1[10],s2[100];
//scanf("%d",&a);
scanf("%s",s1);
n=a; //以下几行是为了确定k的起始位置，为了减少循环。
// printf("(%d)",n);
while((n/=10)!=0) o++;
f=strlen(s1)-o-1;
if(s2[0]==0) f++;
// printf("f=%d|",f);
while(--f)
{
if(f==-1) break;
b*=10;//printf("*");
}
// printf("%d|",b);
//printf("%s",s1);
for(long long k=b;;k++)
{
// int flag=0;
memset(s2,0,sizeof(s2));
sprintf(s2,"%I64d",k*a);
// printf("s2=%s",s2);
p=1; //以下几行是用了uvaoj的all in all那题 判断s1是不是s2的子列的代码
t=-2;
for(i=0;i<strlen(s1)&&p;i++)
{
// printf("*");
for(j=0;j<strlen(s2);j++)
if(s1[i]==s2[j]&&t<j)
{
t=j;
// printf("(s1[%d]=%c,s2[%d]=%c)",i,s1[i],j,s2[j]);
p=0;
break;
}
//printf("%c%c ",s[i],s[j]);
p=!p;
// memset(s,0,sizeof(s));
// memset(t,0,sizeof(t));
//printf("%c%c",s[1],s[1]);
}
if(p) {printf("%I64d\n",k);} //printf("s1=%s,s2=%s,k*a=%d,k=%d,a=%d ",s1,s2,k*a,k,a);break;}
}
}
}
//最终T了，显然会T，就算没T我测了10000， 74562384.得出的是13388961，显然是错误答案，后来发现是a*k爆了int 。
//我用了longlong型，总算没爆了，但是用了十几秒。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long inline ten(int n)
{
long long ans=1;
if(n==0)
return 1ll*1;
for(int i=0;i<n;i++)
ans*=10;
return ans;
}
int main()
{
int a,i;
char c[10];
while(~scanf("%d %s",&a,c)) //好方法，按位非，因为EOF的存储方式是 111...所以按位非后就是0，假。
{
int now=1,nc=0;
for(i=strlen(c)-1;i>=0;i--)
{
nc+=now*(int)(c[i]-'0');
now*=10;
}
int lz=0;
for(i=0;i<strlen(c);i++)
{
if(c[i]!='0')
break;
lz++;
}
long long ab=-1;
long long ans;
int len=strlen(c);
if(lz)
{
len++;
}
int limit=1;
for(;;len++)
{
limit=1;
for(int st=0;st<=len-strlen(c)-(lz>0 ? 1:0);st++)
{
long long base=nc%a;
base*=ten(st);
base%=a;
long long k=ten(strlen(c)+st);
long long ed=ten(len-strlen(c)-st);
for(i=ed/10;i<ed;i++)
{
long long need=a-(((i%a)*k)%a+base)%a;
if(need==a)
need=0;
if(need<limit)
{
long long tmp=i*k+nc*ten(st)+need;
ab=ab==-1 ? tmp : min(ab,tmp);
}
}
limit*=10;
}
if(ab!=-1)
break;
}
ans=ab/a;
printf("%I64d\n",ans);
}
}
// 这是比赛的优秀的代码，运行时间150MS. */
