poj 2325 Persistent Numbers（高精度除法+贪心）

最新推荐文章于 2021-01-22 19:12:00 发布

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本文探讨了如何找出一个数的乘积持久性计算过程中的起始数，即找到一个大数，其数字相乘的第一步结果等于给定的目标数。文章提供了详细的算法思路和实现代码，帮助读者理解这一数学问题的解决方法。

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Persistent Numbers

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

思路：既然后面的数是由前面数的每一位相乘得来的，那么只需要对这个数枚举1~9进行除法，而要想得到最小的数，那么很明显位数越少越好，因此枚举除数时要从9~1，最后反转一下，就得到了最小的前数

代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;

string division(string str,int x)//大整数对小整数的除法
{
    string ans="";
    int len=str.length();
    int y=0;
    for(int i=0; i<len; ++i)
    {
        ans+=((y*10+(str[i]-'0'))/x+'0');
        y=(y*10+(str[i]-'0'))%x;
    }
    while(*(ans.begin())=='0'&&ans.size()>1)
        ans.erase(ans.begin());
    if(y)
        ans=str;
    return ans;
}

int main()
{
    string str;
    while(cin>>str)
    {
        if(str=="-1")
            return 0;
        if(str.length()==1)
            cout<<"1"<<str<<endl;
        else
        {
            string s="",ss=str;
            for(int i=9; i>0; --i)
            {
                while(ss.length()!=1)
                {
                    string qs=division(ss,i);
                    if(qs==ss)
                        break;
                    else
                        s+=(i+'0'),ss=qs;
                }
            }
            if(ss.length()==1)
            {
                s+=ss;
                reverse(s.begin(),s.end());
                cout<<s<<endl;
            }
            else
                cout<<"There is no such number."<<endl;
        }
    }
    return 0;
}