Project Euler #8

I was doing Project Euler #8, which asks:


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?

My solution:

#include 
using namespace std;
#define ull unsigned long long int 
int main() {
// your code goes here
string s;
getline(cin,s);
ull max=0;
for(int i=0;i=max)
   max=prod;
 }
   cout<<max;
   return 0;
}

I have never done this type of question in which we have to read input as a string and than operate on it. I don't know if my approach is right or not. Can anyone point me out where I am wrong?

When analyzing the digits near the end, s[j] will reference an address that is past the end of the string.

You could optimize the solution by taking advantage of the fact that if you have already computed

$$P_0 = a \cdot b \cdot c \cdot d \cdot e \cdot f \cdot g \cdot h \cdot i \cdot j \cdot k \cdot l \cdot m$$

then you could take a shortcut when computing

$$P_1 = b \cdot c \cdot d \cdot e \cdot f \cdot g \cdot h \cdot i \cdot j \cdot k \cdot l \cdot m \cdot n$$

since they share a lot of the same intermediate calculations. In fact,

$$P_1 = \frac{P_0 \cdot n}{a}$$

… as long as \$a \ne 0\$. (You would have to take care to handle zeroes as a special case.)

StackExchange Code Review Q#82976, answer score: 5