Project Euler #8 - Largest product in a series

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

출처: https://projecteuler.net/

주어진 연속된 1000자리 숫자 중에서 연속된 13자리 숫자의 곱 중 가장 큰 수를 구하는 문제입니다.

시간 복잡도를 줄이기위해 기존의 곱셈값에 [i-13] 값을 나누고 [i] 값을 곱해주는 방법을 사용했습니다. 

my solving

c++

#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <string>
using namespace std;
typedef long long ll;
 
 
int main() 
{
    ifstream fin("input.in");
    vector<int> numbers;
 
    string str;
    fin >> str;
 
    for (int i = 0; i < str.size(); i++)
    {
        numbers.push_back(str[i] - '0');
    }
 
    const int adjacentDigits = 13;
 
    ll maxNum = 0;
    ll multi = 1;
    for (int i = 0; i < numbers.size(); i++)
    {
        if (i < adjacentDigits)
        {
            multi *= numbers[i];
            if (i == adjacentDigits - 1)
                maxNum = multi;
            continue;
        }
 
        if (numbers[i - adjacentDigits] == 0)
        {
            multi = 1;
            for (int j = i - adjacentDigits + 1; j <= i; j++)
                multi *= numbers[j];
        }
        else
        {
            multi /= numbers[i - adjacentDigits];
            multi *= numbers[i];
        }
        maxNum = max(maxNum, multi);
    }
 
    cout << maxNum << endl;
 
    system("pause");
    return 0;
}