C++ - 查找数组中第一个和最后一个负数之间的数字

Question

我收到了关于在数组中查找第一个和最后一个负数之间的数字摘要的问题，这是我的代码：

#include <iostream>
using namespace std;

int main(){
float a[] = { 1.5, 55, 5, 4.4,7.1,9,8,2,4.2, -3, -3.2, 5.2, -9 };
float eq = 0;
float sumofnums = 0;

for (int g = 0; g < 13; g++) {

    if (a[g] < 0) {
        sumofnums = a[g];
        cout << sumofnums << endl;
    }
}

我想在第一个和最后一个负数之间找到摘要，我尝试了方法： sumofnums+= a[g];但答案是 -15.2 而实际上是 -10。

Answer 1

一种可能的解决方案：

#include <iostream>
using namespace std;

int main() {
  float a[] = { 1.5, 55, 5, 4.4,7.1,9,8,2,4.2, -3, -3.2, 5.2, -9 };
  float sumofnums = 0;

  int size = sizeof(a) / sizeof(a[0]);  // size is 13 here, this allows you
                                        // to determine the size automatically
  int first = 0;
  int last = size - 1;

  for (int g = 0; g < size; g++) {
    if (a[g] < 0) {
      first = g;
      break;
    }     
  }

  for (int g = size - 1; g >= 0; g++) {
    if (a[g] < 0) {
      last = g;
      break;
    }
  }

  for (int g = first; g <= last; g++) {
    sumofnums += a[g];
  }

  cout << "Sum is " << sumofnums << endl;
}

不用cmets应该就够清楚了。

Answer 2

这个循环

for (int g = 0; g < 13; g++) {

    if (a[g] < 0) {
        sumofnums = a[g];
        cout << sumofnums << endl;
    }
}

依次将数组的负值赋给变量sumofnums，而不是累加所需范围内的所有值。

如果使用标准算法，那么程序可能看起来像

#include <iostream>
#include <iterator>
#include <algorithm>
#include <numeric>
#include <functional>

int main()
{
    float a[] = { 1.5, 55, 5, 4.4, 7.1, 9, 8, 2, 4.2, -3, -3.2, 5.2, -9 };
    float sumofnums = 0.0f;

    auto first = std::find_if(std::begin(a), std::end(a),
        std::bind2nd(std::less<float>(), 0.0f));

    if (first != std::end(a))
    {
        auto last = std::find_if(std::rbegin(a), std::rend(a),
            std::bind2nd(std::less<float>(), 0.0f)).base();

        sumofnums = std::accumulate(first, last, sumofnums);
    }

    std::cout << sumofnums << std::endl;

    return 0;
}

它的输出是

-10

如果使用手动编写的循环，那么程序可能看起来像

#include <iostream>

int main()
{
    float a[] = { 1.5, 55, 5, 4.4, 7.1, 9, 8, 2, 4.2, -3, -3.2, 5.2, -9 };
    const size_t N = sizeof(a) / sizeof(*a);
    float sumofnums = 0.0f;

    const float *first = a;

    while (not (first == a + N) and not (*first < 0.0f))
    {
        ++first;
    }

    if (first != a + N )
    {
        const float *last = a + N;

        while (not (last == a) and not (*(last - 1 ) < 0.0f))
        {
            --last;
        }

        for (; first != last; ++first) sumofnums += *first;
    }

    std::cout << sumofnums << std::endl;

    return 0;
}

输出与上图相同。