Problem 8:Largest product in a series

题目：
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?

个人认为本人的方法时空复杂度已经降至最低，代码不一定好懂，但是绝对是对的。

Answer:23514624000

#include<iostream>
#include<cstdio>
using namespace std;
int a[14];//记录13个位置的数字
int main()
{
    int t = 0, temp, s = 1;//t代表已经记录的位置数，s代表即将被更改的位置
    bool flag = false;
    long long sum = 0LL, now = 1LL;
    int cnt = 0;
    while(++cnt <= 1000)
    {
        scanf("%1d", &temp);
        if(0 == temp)
        {
            t = 0;
            s = 1;
            flag = false;
            continue;
        }
        if(t < 13)
            a[++t] = temp;
        else if(flag)
        {
            now = now / a[s] * temp;
            if(now > sum)
                sum = now;
            a[s] = temp;
            s = s % 13 + 1;
        }
        if(13 == t && !flag)
        {
            now = 1LL;
            for(int k = 1; k <= 13; k++)
                now *= a[k];
            if(now > sum)
                sum = now;
            flag = true;
        }
    }
    printf("%I64d\n", sum);
    return 0;
}