2.13 Maximum multiplication of sub-arrays

Topic:

Given a shaped array of length n, only multiplication is allowed and no division can be used. Calculates the largest group of products in a combination of any number of N-1.

Method One:

#include <iostream> #define MAXN 10000using namespace Std;int N, A[maxn], S[MAXN], T[MAXN], p[maxn];//s[i] The product of the first element of the array//t[i] represents the product of N-i elements after the array//p[i] = s[i-1] * t[i+1];   P[i] indicates that the array, in addition to the first element, is the product of the other N-1 elements//s[i]=a[0]*a[1]* ... A[i], t[i]=a[i]*a[i+1]* ... a[n-1]//P[i] = s[i-1] * t[i+1]. int main () {cin >> n;for (int i = 0; i < n; ++i) cin >> A[i];int SV = 1, TV = 1;for (int i = 0; i < n; ++i ) {Int J = n-i-1;sv *= a[i];tv *= a[j];s[i] = sv;t[j] = TV;} int maxnum = p[0] = t[1];    P[0] and t[1] represent the values that are multiplied by the remaining N-1 elements except for the first element. for (int i = 1; i < n; ++i) {p[i] = s[i-1] * t[i+1];maxnum = Maxnum < P[i]? P[i]: maxnum;} cout << maxnum << endl;return 0;}

Method Two:

Ideas:

The maximal product problem of Subarray can be divided into the following situations.

1. There are more than one zero in the array, the maximum product is 0;

2. There is only one 0 in the array, and an odd number of negative numbers, the maximum product must be 0;

3. There is only one 0 in the array, and an even number of negative numbers, the maximum product is the product of the element that removes 0;

4. There is no zero in the array, and there are odd numbers of negative numbers, then the maximum product is the product of the negative number minus the absolute value;

5. There is no zero in the array, and an even number of negative numbers, the maximum product is the product of the lowest positive number removed.

So the number of positive numbers in the array, the number of negative numbers, and the number of 0! This is the simplest way!

2.13 Maximum multiplication of sub-arrays