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 = 25;
LL dp_max[maxn],dp_min[maxn];
int iCase = 1;

int main()
{
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    int n;
    while(cin >> n){
        dp_max[0] = dp_min[0] = 0LL;
        LL Max = 0;
        int x;
        for(int i = 1;i <= n;i++){
            cin >> x;
            if(x > 0){
                dp_max[i] = dp_max[i-1]*x;
                dp_min[i] = dp_min[i-1]*x;
                if(!dp_max[i]) dp_max[i] = x;
            }
            else if(x < 0){
               dp_max[i] = dp_min[i-1]*x;
               dp_min[i] = dp_max[i-1]*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 0;
}