哈理工OJ 2116 Maximum continuous product（思维的体操）

Maximum continuous product
Time Limit: 1000 MS Memory Limit: 32768 K
Total Submit: 111(37 users) Total Accepted: 34(30 users) Rating: Special Judge: No
Description
Wind and his GF(game friend) are playing a small game. They use the computer to randomly generated a number sequence which only include number 2,0 and -2. To be the winner,wind must

have to find a continuous subsequence whose continuous product is maximum.

For example, we have a sequence blow:

2 2 0 -2 0 2 2 -2 -2 0

Among all of it’s continuous subsequences, 2 2 -2 -2 own the maximum continuous product.

(2*2*(-2)*(-2) = 16 ,and 16 is the maximum continuous product)

You,wind’s friend,can give him a hand.

Input
The first line is an integer T which is the Case number(T <= 200).

For each test case, there is an integer N indicating the length of the number sequence.（1<= N <= 10000)

The next line,there are N integers which only include 2,0 and -2.

Output
For each case,you have to output the case number first(Reference the sample).

If the answer is smaller than 0, you just need to output 0 as the answer.

If the answer’s format is 2^x,you need to output the x as the answer.

Output the answer in one line.

Sample Input
2
2
-2 0
10
2 2 0 -2 0 2 2 -2 -2 0
Sample Output
Case #1: 0
Case #2: 4

题目大意：给你一组数，这组数由0，-2，2组成。让你求出最大的连续积，然后输出它是2的多少次方，如果最大积为0 的话输出0。

思路：暴力扫。

下面是AC代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[10005];int main(){    int t,iCase=0;    scanf("%d",&t);    while(t--)    {        iCase++;        int n;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        int maxn=0,cnt,flag;        for(int i=0;i<n;i++)        {            if(a[i]==0)            {                continue;            }            cnt=0;            flag=0;            for(int j=i;j<n;j++)            {                if(a[j]==0)                {                    break;                }                if(a[j]==2)                {                    cnt++;                }                if(a[j]==-2)                {                    flag=1-flag;                    cnt++;                }                if(flag==0)                {                    maxn=max(maxn,cnt);                }            }        }        printf("Case #%d: %d\n",iCase,maxn);    }    return 0;}