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.

对于每个数，都设置以此数结尾的最大正数pos和最小负数neg，

当data[i]是正数的时候，pos=pos*data[i]，neg=neg*data[i]

当data[i]是0的时候，pos=neg=0;

当data[i]是负数的时候，neg*data[i]为最大正数，pos*data[i]就是最小负数，

每次循环更新 maxn；

注意数据范围 用long long

#include <iostream>#include<cstdio>#include<cstring>using namespace std;int data[20];int main(){    int n,num=0;    while(scanf("%d",&n)!=EOF && n)    {        for(int i=1;i<=n;i++)        scanf("%d",&data[i]);        long long maxn=0,neg=0,pos=0;        for(int i=1;i<=n;i++)        {            if(data[i]>0)            {                pos=pos*data[i];                neg=neg*data[i];                if(pos==0)                    pos=data[i];            }            if(data[i]==0)                pos=neg=0;            if(data[i]<0)            {                long long temp=pos*data[i];                pos=neg*data[i];                neg=temp;                if(neg==0)                    neg=data[i];            }            if(maxn<pos)                maxn=pos;        }        printf("Case #%d: The maximum product is %lld.\n\n",++num,maxn);    }    return 0;}