UVA OJ-11095 Maximum Product (Brute force solution method)

Given a sequence of integers S = {S1, S2, ..., Sn}, you should determine what is the value of the maximum Positiv E product involving consecutive terms of S. If you cannot find a positive sequence, you should consider 0 as the value of the maximum product.

Input

Each test case is starts with 1≤n≤18, and the number of elements in a sequence. Each element Si is a integer such that -10≤si≤10. Next Line would has N integers, representing the value of each element in the sequence. There is a blank line after each test case. The input is terminated by end of file (EOF).

Output

For each test case you must print the message: Case #M: The maximum product is P., where M is the number Of the test case, starting from 1, and P is the value of the maximum product. After each test, you must print a blank line.

Sample Input

32 4-352 5-1 2-1

Sample Output

Case #1: The maximum product was 8.Case #2: The maximum product is 20.
Main topic: Product of maximal continuous sub-series

#include <iostream>using namespace Std;int main () {  int n, a[20], count = 1;  while (CIN >> N) {for    (int i = 0; i < n; i++)      cin >> A[i];    Long Long max = 0;//If the maximum product is not a positive number, output 0 for     (int i = 0; i < n; i++) {//enumerate from the beginning       long long sum = 1;      for (int j = i; J < N; j + +) {//enumeration end         sum *= a[j];        if (Sum > Max)          max = sum;      }     }      cout << "Case #" << count++ << ": The maximum product is" << Max << "." << endl;
   
    cout << Endl;  
   

　　

UVA OJ-11095 Maximum Product (Brute force solution method)