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
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 ;
}