最大乘积(Maximum Product,UVA 11059)

Problem D - Maximum Product

Time Limit: 1 second

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

Input

Each test case starts with 1 ≤ N ≤ 18, the number of elements in a sequence. Each elementS_i is an integer such that-10 ≤ S_i ≤ 10. Next line will haveN 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., whereM is the number of the test case, starting from1, andP is the value of the maximum product. After each test case you must print a blank line.

Sample Input

32 4 -352 5 -1 2 -1

Sample Output

Case #1: The maximum product is 8.Case #2: The maximum product is 20.

#include <stdio.h>#include <set>using namespace std;int main(){int n;int * val = NULL;set<long long> s;while((scanf("%d",&n) == 1) && n != 0){val =  new int[n];int i = 0;for (;i  < n; i++) {scanf("%d",&val[i]);}s.clear();int j ;for(i = 0;i < n -1;i++){//枚举起点for (j = i; j < n; ++j) {//枚举终点int k;long long ji = 1;for(k = i; k <= j;k++){ji *= val[k];}s.insert(ji);}}long long max = *(s.rbegin());if(max > 0){printf("%ld\n",max);}else{printf("0\n");}                delete val;        }return 0;}