HDUOJAlexandra and A*B Problem

/*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. */