Maximum Product UVA - 11059（枚举入门）

Given a sequence of integers

S

=

f

S

1

;S

2

;:::;S

n

g

, you should determine what is the value of the

maximum positive product involving consecutive terms of

S

. If you cannot nd 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

S

i

is

an integer such that

 

这道题只需要枚举起点到终点的连续序列就行。但是有个坑点：当m初始化为-11时，m<=0时要输出0，千万不能把=号忽略了（m代表输出的结果）；只不过有一个解决办法，就是m初始化的时候直接m=0.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <sstream>
#include <cmath>
#include <queue>
using namespace std;
const int maxn = 100005;
int a[maxn];

int main()
{
    int n,kase = 1;
    while(cin>>n){
        long long t = 1,m = -11;
        memset(a,0,sizeof(a));
        for(int i = 0;i < n;i++)
            cin>>a[i];
        for(int i = 0;i < n;i++){
            for(int j = i;j < n;j++){
                for(int k = i;k <= j;k++){
                    t *= a[k];
                }
                //cout<<"t: "<<t<<endl;
                if(t > m) m = t;
                t = 1;
            }
        }
        cout<<"Case #"<<kase++<<": The maximum product is ";
        if(m <= 0) cout<<0<<"."<<endl;
        else
            cout<<m<<"."<<endl;
        cout<<endl;
    }
    return 0;
}