UVA-11059 Maximum Product

Given a sequence of integers S = {S1,S2,…,Sn}, you should determine what is the value of the maximum positive 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 starts with 1 ≤ N ≤ 18, the number of elements in a sequence. Each element Si is an integer such that −10 ≤ Si ≤ 10. Next line will have 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 case you must print a blank line.

Sample Input

3

2 4 -3

5

2 5 -1 2 -1

Sample Output

Case #1: The maximum product is 8.

Case #2: The maximum product is 20.

分析：枚举开头和结尾，暴力求解。注意用long long保存乘积，|product|<=10^18

Source:

#include<stdio.h>
int a[20];
int main()
{
int m=0,n;
int i,j;
long long p,max;
while(scanf("%d",&n)!=EOF)
{
m++;
max=0;
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
{
p=1;
for(j=i;j<n;j++)
{
p*=a[j];
if(p>max)
max=p;
}
}
printf("Case #%d: The maximum product is %lld.\n\n",m,max);
}
return 0;
}