Hdu 4561 Simulation

Continuous Maximum Product
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission (s): 699 Accepted Submission (s): 275

 

Problem Description
Tom and his good friend Tom are playing a game. The computer randomly generates an array consisting of-2, 0, and 2, who first calculates the maximum product of a continuous element in this array, even if it wins!

For example, we have the following random array:
2 2 0-2 0 2-2-2 0
Among the many consecutive subsequences in this array, the product of the continuous subsequence 2-2-2 is the largest.

James asks you to calculate the maximum value.
 

Input
Enter a positive integer T in the first line, indicating a total of T groups of data (T <= 200 ).
For the next T group data, input N in the first row of each group data, indicating the total number of elements in the array (1 <= N <= 10000 ).
Then, enter N elements consisting of 0,-2, and separate them with spaces.
 

Output
For each group of data, the number of cases is output first.
If the final answer is less than or equal to 0, 0 is output directly.
Otherwise, if the answer is 2 ^ x, output x.
Each group of data occupies one row. For the specific output format, see the sample.
 

Sample Input
2
2
-2 0
10
2 2 0-2 0 2-2-2 0

Sample Output
Case #1: 0
Case #2: 4

Source
2013 Jinshan xishanju creative game program challenge-semi-finals (2)
 

Recommend
Liuyiding
 
Consider 0 as a partition, divide the number of strings into multiple segments, and then calculate the continuous product of a maximum of several numbers in each small segment as a positive number, and then take a maximum value.
 

#include<stdio.h>int a[50000+100],n;int solve(int s,int e){    int i,cnt=0,st,ed,flag=1,ans=0;    if(s==n+1) return 0;     for(i=s;i<e;i++)     {         if(a[i]==-2)         {             if(flag)  {flag=0;st=i;}             ed=i;             cnt++;         }     }    if(cnt%2==0) return e-s;    if(ans<st-s) ans=st-s;    if(ans<e-st-1) ans=e-st-1;    if(ans<e-ed-1) ans=e-ed-1;    if(ans<ed-s)  ans=ed-s;    return ans;}int main(){    int cas,i,k,ans;    scanf("%d",&cas);    for(k=1;k<=cas;k++)    {        ans=0;        scanf("%d",&n);        for(i=1;i<=n;i++)  scanf("%d",&a[i]);        a[0]=0;        for(i=0;i<=n;)        {            if(a[i]==0)            {               int ii=i+1;                while(1)                {                    i++;                    if(i==n+1||a[i]==0)                        {                             int mid=solve(ii,i);                             if(ans<mid)  ans=mid;                              break;                        }                }            }        }        printf("Case #%d: %d\n",k,ans);    }   return 0;}