hdu 5109 Alexandra and A*B Problem（取模，枚举，数学数论）

最新推荐文章于 2020-08-04 23:38:19 发布

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#ACM #HDU #数学

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本文解析了一道编程题：给定A和字符串S，找到最小正整数B，使得A*B的结果中包含S作为子串。文章详细阐述了解题思路，采用取模技巧缩小搜索范围，最终通过枚举得出答案。

Alexandra and A*B Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 804 Accepted Submission(s): 212

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

Sample Output

题意：A*B=T，S是T的字串，比如T是11000 S可能是100 1100...

给你A和S，找出最小的B满足条件

思路：刚看这题完全没思路...还是取模的题目做少了。

我们假设T=XsY X和Y分别是补在s前面和后面的一个串，设len为s的长度，len1为Y的长度

则T=（X*10^len+s）*10^len1+Y 由于A*B=T,所以有T%A==0 所以对于大于a的X和Y我们就可以通过取模减小0范围

这样X~[0,a) 当然如果s有前导0，X~[1,a)，因为0相当于什么也不加。

当s==0的时候，直接输出0 任何数乘以0都为0

然后我们只要枚举X和len1就可以得到Y，然后就可以求出满足条件的最小的T

然后T/a即为最后的答案

注意因为Y~[0,a）所以len1范围为0~4 Y不一定是一个整数，可以形如0083

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
long long pow_(long long a,long long n)
{
    long long ans=1;
    while(n)
    {
        if(n&1) ans=ans*a;
        a=a*a;
        n>>=1;
    }
    return ans;
}
int main()
{
    long long a;
    char s[10];
    while(~scanf("%lld",&a))
    {
        scanf("%s",s);
        long long num=0,len=strlen(s);
        for(int i=0; i<len; i++)
            num=num*10+s[i]-'0';
        int flag=0;
        if(len>1&&s[0]=='0') flag=1;
        if(len==1&&num==0) printf("0\n");
        else
        {
            long long i=0,last_ans=-1;
            if(flag) i=1;
            for(; i<a; i++)
            {
                long long ans=i*pow_(10,len)+num;
                for(long long j=1; j<=10000; j*=10)
                {
                    long long mod=ans*j;
                    long long y=(a-mod%a)%a;
                    if(y<j)
                    {
                        mod+=y;
                        if(last_ans==-1) last_ans=mod;
                        else last_ans=min(last_ans,mod);
                    }
                }
            }
            printf("%lld\n",last_ans/a);
        }
    }
    return 0;
}