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?

1.滑动窗口法：

  1. 滑动窗口法：使用for循环模拟窗口移动
  2. 窗口最后一个数字不为0便将新数字乘入乘积,否则0的计数器++
  3. 窗口最前面的数字不为0便将最前面的数字除掉，否则0的计数器–
  4. 当0的计数器为零时，窗口乘积有效可尝试更新答案

注意：由于0乘除运算的特殊性，在滑动窗口法的基础上，还需要额外定义一个关于0的计数器

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#include<iostream>
using namespace std;

char str[1005];

int main(){
    long long ans = 0, now = 1;
    int zero_cnt = 0;
    
    cin >> str;
    //1.利用for循环模拟窗口向后移动，当循环遇到字符串str[i]结束符\0时将结束循环
    for(int i = 0; str[i]; ++i){
        if(i < 13){
            //前13个数字中没有0，可以直接乘入窗口乘积
            now *= str[i] - '0';
        }else{
            //2.窗口最后一个数字不为0便将新数字乘入乘积,否则计数器++
            if(str[i] != '0'){
            	//(1)前13个数字之后不为0的数字:窗口乘积乘入最后面的数字
            	now *= str[i] - '0';
        	}else{
            	//(2)前13个数字之后为0:计数器记录加1
            	zero_cnt++;
        	}
            
            //3.窗口最前面的数字不为0便将最前面的数字除掉，否则计数器--
            if(str[i - 13] != '0'){
                //(1)前13个数字之后不为0的数字:窗口乘积除掉最前面的数字
                now /= str[i - 13] - '0';
            }else{
                //(2)前13个数字之后为0:计数器记录减1
                zero_cnt--;
            }
        }
        
        //4.动态更新答案
        if(zero_cnt == 0){
            ans = max(ans, now);
        }
    }
    
    cout << ans << endl;
    return 0;
}

补充：滑动窗口分为动态窗口（双指针法） 与 静态窗口