数组子串的最大乘积

“==============================================”一下的文章是错误的，因为乘积和求和不同，求和过程中正负属性会更改，而乘积不会。所以对于{-10，-10，1}这样的例子，那个算法是错误的。在网上看到一个DP算法，实现了一下。如下：
假设b[i]以a[i]结尾的最大连续子串乘积，d[i]为最小。则
b[0]=d[0]=a[0]
b[i] = max{b[i-1]*a[i], d[i-1]*a[i], a[i]};
d[i] = min{b[i-1]*a[i], d[i-1]*a[i], a[i]};
而整个问题的解就是max{b[i]}

#define __max__(a,b) ((a)>(b)?(a):(b))
#define __min__(a,b) ((a)<(b)?(a):(b))
double maxmulti2(int *p, int len)
{
    double maxend, minend, maxres;
    maxend = minend = maxres = *p;
    for (int i = 1; i < len; ++i)
    {
        double end1 = maxend* *(p + i);
        double end2 = minend* *(p + i);

        maxend = __max__(__max__(end1, end2), *(p + i));
        minend = __min__(__min__(end1, end2), *(p + i));
        maxres = __max__(maxend, maxres);
    }
    return maxres;
}

==============================================
和数组子串的最大和 类似，数组子串的最大乘积也有两种求值方法。
对应的，之前的判断和改为判断积；之前的跟‘0’比较变成跟‘1’比较（若B>1，则AB比A大，否则比A小），对应代码为：

#include <iostream>
using namespace std;

double d[7];
double End[7];

double maxsub(double *p, int len)
{
    if (p == NULL || len <= 0)
        return -1;

    d[0] = *p;
    double sum, tmpsum;
    sum = tmpsum = p[0];
    for (int i = 1; i < len; ++i)
    {
        if (tmpsum>1)
            tmpsum *= (*(p + i));
        else
            tmpsum = (*(p + i));
        if (tmpsum > sum)
            sum = tmpsum;
    }
    return sum;

}

double max(double a, double b)
{
    return a > b ? a : b;
}
double maxsub2(double *p, int len)
{
    if (p == NULL || len <= 0)
        return -1;

    d[0] = End[0] = *p;
    for (int i = 1; i < len; ++i)
    {
        End[i] = max(End[i - 1] * p[i], p[i]);
        d[i] = max(End[i], d[i - 1]);
    }
    return d[len - 1];
}
int main()
{
    double a[7] = { -2.5, 4, 0, 3, 0.5, 8, -1 };
    cout << maxsub(a, sizeof(a) / sizeof(double)) << endl;
    cout << maxsub2(a, sizeof(a) / sizeof(double)) << endl;

    system("pause");


    return 0;
}