Problem 8:Largest product in a series

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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;}

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