Hust oj 2116 Maximum continuous product（水题）-优快云博客

本文链接：https://blog.youkuaiyun.com/Sara_YF/article/details/52075757

本文介绍了一道编程题目，任务是找出由2、0和-2组成的数列中乘积最大的连续子序列，并返回该最大乘积的指数形式。文章提供了完整的C++代码实现，包括输入测试用例和输出结果。

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Maximum continuous product

Time Limit: 1000 MS

Memory Limit: 32768 K

Total Submit: 119(40 users)

Total Accepted: 38(33 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

题意有点坑，没明说可以在一段中截取一段计算乘积

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;

const int maxn = 10005;
const int Inf = 0x3f3f3f;
int a[maxn];
int t,n;

int main()
{
    int ans = 0;
    scanf("%d",&t);
    while(t--)
    {
        int cnt1 = 0;
        int cnt2 = 0;
        int cnt3 = 0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        int Max = 0;
        for(int i=0;i<n;i++)
        {
           if(a[i] == 0) continue;
           int cnt = 0;
           int flag = 0;
           for(int j=i;j<n;j++)
           {
               if(a[j] == 0) break;
                if(a[j] == 2) cnt++;
                if(a[j] == -2)
                {
                    flag = !flag;
                    cnt++;
                }
                if(!flag)
                Max = max(cnt,Max);
         }
        }
        printf("Case #%d: ",++ans);
        printf("%d\n",Max);
    }
}