L1-046 divide singles (20 points)

The so-called “ A bachelor ”, It's not about being single ~ It's all about 1 The numbers that make up , such as 1、11、111、1111 etc. . It is said that any single can be denied by one 5 Odd division at the end . such as ,111111 Can be 13 to be divisible by . Now? , Your program will read in an integer x, The whole number must be odd and it doesn't have to be 5 ending . then , After calculation , Output two numbers ： The first number s, Express x multiply s A bachelor , The second number n It's the number of singles . Of course, such a solution is not the only one , The question requires you to output the smallest solution .

Tips ： One obvious way is to gradually increase the number of singles , Until it's divisible x until . But the difficulty is ,s It could be a very large number —— such as , Program input 31, Then output 3584229390681 and 15, because 31 multiply 3584229390681 The result is 111111111111111, altogether 15 individual 1.

Input format ：

Type in a line to give a not to 5 A positive odd number at the end x（<1000）.

Output format ：

Output the corresponding smallest... In one line s and n, In the meantime 1 Space separation .

sample input ：

31

sample output ：

3584229390681 15

notes ：

Deformation of high precision multiplication

Code ：

#include <iostream>
#include <vector>

using namespace std;

vector<int> mul(vector<int> & a, int b){
    vector<int> c;
    int t = -1;
    while(t){
        if(t == -1) t = 0;
        for(int i = 0; i <= 9; i ++)
            if((i * b + t) % 10 == 1){
                a.push_back(i);
                c.push_back(1);
                t += i * b;
                t /= 10;
                break;
            }
    }
    while(a.size() > 1 && a.back() == 0) a.pop_back();
    return c;
}

int main(){
    int b;
    cin >> b;

    vector<int> a;
    vector<int> c = mul(a, b);

    for(int i = a.size() - 1; i >= 0; i --) 
        cout << a[i];
    cout << " " << c.size();
}