UVA11059-Maximum Product(动态规划)

Problem UVA11059-Maximum Product

Accept：4769  Submit：38713

Time Limit: 3000 mSec

 Problem Description

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 Ouput

Case #1: The maximum product is 8.

Case #2: The maximum product is 20.

题解：lrj给的是暴力的做法，不够这个题显然可以dp去做，维护两个dp数组，一个维护最大正值，一个维护最小负值，状态转移很简单，看代码就能明白。

P.S.原来之一用printf("%lld\n",x)来输出long long,这一次用了I64d输出，WA到怀疑人生，不知道为啥。（写这一篇题解就是为了记录这个WA点）

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 using namespace std;
 typedef long long LL;

 const int maxn = ;
 LL dp_max[maxn],dp_min[maxn];
 int iCase = ;

 int main()
 {
     //freopen("input.txt","r",stdin);
     //freopen("output.txt","w",stdout);
     int n;
     while(cin >> n){
         dp_max[] = dp_min[] = 0LL;
         LL Max = ;
         int x;
         for(int i = ;i <= n;i++){
             cin >> x;
             if(x > ){
                 dp_max[i] = dp_max[i-]*x;
                 dp_min[i] = dp_min[i-]*x;
                 if(!dp_max[i]) dp_max[i] = x;
             }
             else if(x < ){
                dp_max[i] = dp_min[i-]*x;
                dp_min[i] = dp_max[i-]*x;
                if(!dp_min[i]) dp_min[i] = x;
             }
             else{
                 dp_max[i] = dp_min[i] = 0LL;
             }
             if(Max < dp_max[i] && x) Max = dp_max[i];
         }
         printf("Case #%d: The maximum product is %lld.\n\n",iCase++,Max);
     }
     return ;
 }