hrbust/哈理工oj 2116 Maximum continuous product【水题】

Maximum continuous product

Time Limit: 1000 MS Memory Limit: 32768 K

Total Submit: 107(37 users) Total Accepted: 33(29 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

Source

ACM-ICPC黑龙江省第九届大学生程序设计竞赛选拔赛(2)

竟然直接暴力就能过T___T害我wa了好久T-T

AC代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int a[100000];
int main()
{
int t;
int kase=0;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
int output=0;
int cnt=0;
int flag=0;
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&&cnt>output){output=cnt;}
}
}
printf("Case #%d: %d\n",++kase,output);
}
}
/*
100
4
-2 2 2 2
4
-2 2 0 2
4
2 -2 2 0
8
2 -2 2 0 2 -2 2 2
10
2 -2 -2 2 2 2 -2 2 0 -2
10
0 2 2 -2 2 0 -2 0 0 0
8
2 0 2 -2 2 2 -2 2
*/