UVA11059 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 Siisan integer such that −10 ≤ Si ≤ 10. Next line will have N integers, representing the value of eachelement in the sequence.There is a blank line after each test case. The input is terminated by end offile (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.

UVA的格式是真的牛掰！

看我加粗的两行。

一开始我以为是输入完以后输出一个空行。

在输出数据后再输出一个空行。

结果是只要后面那个，QAQ！

不多说，来说思路：

求连续子串的积，我们可以找到起点和终点即可，又因为每个元素的绝对值不超过10且只有18个元素，最大枚举数不超过10的18次方，我们可以用 long long存储即可。

#include<iostream>#include<cstdio>using namespace std;int main() {int n;int T=0;while(scanf("%d",&n)!=EOF){int a[20];long long max=0,sum=1;for(int i=0;i<n;i++){cin>>a[i];}//if(T++) cout<<"\n";T++;for(int i=0;i<n;i++){  for(int j=i;j<n;j++)  {  sum*=a[j];  if(sum>max)    max=sum;  }  sum=1;}  if(max<=0){   printf("Case #%d: The maximum product is %ld.\n\n",T,0);}  else{   printf("Case #%d: The maximum product is %ld.\n\n",T,max);}      }return 0;}