hdu 5109(构造数+对取模的理解程度)

Alexandra and A*B Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 802    Accepted Submission(s): 211

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

 

题意:给出一个数字a,以及一个串s(lens<=8)找到一个最小的数字 b ,使得 s 是 a*b = t 的一个连续子序列.

题解:非常巧妙的题目。假设t = xsy , 那么我们可以写出表达式 t = (x*10^lens+s)*10^len+y ,又因为 t%a == 0 所以对于最小的 t 式子的每一部分,都可以对 a 取模,

所以我们可以知道 x<a ，因为如果 x >= a，我们可以通过取模操作让 x 变小, 然后如果s[0]==0,那么x的下限就是1，否则x的下限为 0，然后对于 10^len 我们也可以通过同样的道理知道 10^len<= a <= 10^4 ,然后我们化出：

设 k = (x*10^lens+s)*10^len

所以 (k+y)%a=0

y = (-k%a+a)%a (y<10^len)

所以通过枚举 x,10^len 得到最终的答案。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long LL;

int main()
{
    LL a;
    char s[];
    while(scanf("%lld%s",&a,s)!=EOF){
        int len = strlen(s);
        if(strcmp(s,"")==){
            printf("0\n");
            continue;
        }
        LL ls = ,mid=;
        for(int i=;i<len;i++) ls*=;
        for(int i=;i<len;i++){
            mid = mid*+s[i]-'';
        }
        LL ans = -;
        for(LL i=;i<=;i*=){
            for(LL j=(s[]=='');j<a;j++){
                LL t = (j*ls+mid)*i;
                LL y = (a-t%a)%a;
                if(y>=i) continue;
                if(ans<) ans = t+y;
                else ans = min(t+y,ans);
            }
        }
        printf("%I64d\n",ans/a);
    }
    return ;
}