UVA11059 - Maximum Product

Maximum Product

Time limit: 3.000 seconds

Given a sequence of integers S = {S1, S2, . . . , Sn}, you should determine what is the value of themaximum 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.’, whereM is the number of the test case, starting from 1, and P is the value of the maximum product. Aftereach 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.

题意是在n个元素组成的序列S中，找一个乘积最大的连续子序列。一个枚举的水题，直接枚举起点和终点即可！（需注意乘积为负数时输出0）

#include <stdio.h>#include <iostream>#include <string.h>#include <algorithm>using namespace std;const int Max = 20;const int INF = 1000000000;int a[Max], n;int main(){    int temp = 0;    while(~scanf("%d",&n))    {        memset(a,0,sizeof(a));        for(int i = 0; i < n; i++)            scanf("%d",&a[i]);        long long sum, ans = 0;        for(int i = 0; i < n; i++)        {            for(int j = i; j < n; j++)            {                sum = 1;                for(int k = i; k <= j; k++)                    sum *= a[k];                ans = max(ans,sum);            }        }        printf("Case #%d: The maximum product is %lld.\n\n",++temp,ans);    }    return 0;}