C++程序最小乘积子集

给定一个数组a，我们必须找到该数组中元素子集的最小乘积。最小乘积也可以是单个元素。

例子：

输入: a[] = { -1, -1, -2, 4, 3 }
输出: -24
解释: 最小乘积为( -2 * -1 * -1 * 4 * 3 ) = -24

输入: a[] = { -1, 0 }
输出: -1
解释: -1(单个元素)是最小乘积

输入: a[] = { 0, 0, 0 }
输出: 0

一个简单的解决方案是生成所有子集，找到每个子集的乘积并返回最小乘积。

更好的解决方案是使用以下事实。

如果负数的数量为偶数且没有零，则结果是减去所有负数中的最大值。
如果负数的数量为奇数且没有零，则结果就是所有数字的乘积。
如果有零且所有数字都是正数，则结果为0。当没有负数并且所有其他数字都是正数时，我们的结果应为第一个最小的正数。

// CPP program to find maximum product of
      // a subset.
      #include <bits/stdc++.h>
      using namespace std;

      int minProductSubset(int a[], int n)
      {
          if (n == 1)
              return a[0];

          // Find count of negative numbers, count
          // of zeros, maximum valued negative number,
          // minimum valued positive number and product
          // of non-zero numbers
          int max_neg = INT_MIN;
          int min_pos = INT_MAX;
          int count_neg = 0, count_zero = 0;
          int prod = 1;
          for (int i = 0; i < n; i++) {

              // If number is 0, we don't
              // multiply it with product.
              if (a[i] == 0) {
                  count_zero++;
                  continue;
              }

              // Count negatives and keep
              // track of maximum valued negative.
              if (a[i] < 0) {
                  count_neg++;
                  max_neg = max(max_neg, a[i]);
              }

              // Track minimum positive
              // number of array
              if (a[i] > 0)
                  min_pos = min(min_pos, a[i]);

              prod = prod * a[i];
          }

          // If there are all zeros
          // or no negative number present
          if (count_zero == n
              || (count_neg == 0 && count_zero > 0))
              return 0;

          // If there are all positive
          if (count_neg == 0)
              return min_pos;

          // If there are even number of
          // negative numbers and count_neg not 0
          if (!(count_neg & 1) && count_neg != 0) {

              // Otherwise result is product of
              // all non-zeros divided by maximum
              // valued negative.
              prod = prod / max_neg;
          }

          return prod;
      }

      int main()
      {
          int a[] = { -1, -1, -2, 4, 3 };
          int n = sizeof(a) / sizeof(a[0]);
          cout << minProductSubset(a, n);
          return 0;
      }