Description
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.
题目是求最大连续子串乘积;
枚举起点和终点即可。。。。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 7 const int Max = 20; 8 const int INF = 1000000000; 9 int a[Max], n; 10 11 int main() 12 { 13 int temp = 0; 14 while(~scanf("%d",&n)) 15 { 16 memset(a,0,sizeof(a)); 17 for(int i = 0; i < n; i++) 18 scanf("%d",&a[i]); 19 long long sum, ans = 0; 20 for(int i = 0; i < n; i++) 21 { 22 for(int j = i; j < n; j++) 23 { 24 sum = 1; 25 for(int k = i; k <= j; k++) 26 sum *= a[k]; 27 ans = max(ans,sum); 28 } 29 } 30 printf("Case #%d: The maximum product is %lld. ",++temp,ans); 31 } 32 return 0; 33 }