如何计算总和的总和和平均值，用户输入

Question

循环？

#include <iostream>
using namespace std;
#include <iomanip>

int main(int argc, const char * argv[]) {
    int x;
    double avg = 0.0;
    int count = 0;
    int sum = 0;
    
    // ask user for input
    cout << ("Welcome to the greatest calculator!\n");
    cout << ("Please enter 10 integers seperated by spaces \n");
    
    do {
        std::cin >> x;
        sum = sum + x;
        count = count + 1;
    }
    while (count < 10);
    
    // calculate average
    avg = sum/10.0;
    
    // output average
    cout << fixed;
    cout << "For all 10 numbers the sum is " << sum << "." "The average is " << setprecision (2) << sum/10.0 << ".\n";
    return 0;
}

输出应该看起来像这样。

Please enter 10 integers separated by spaces:
1 -1 45 17 28 -2 0 9 -14 11

Upon our intelligent calculations, here is the result:
+ There are 7 positive numbers, sum = 111.00 and average = 15.86
+ There are 3 negative numbers, sum = -17.00 and average = -5.67
+ For all 10 numbers, sum = 94.00 and average = 9.40

您可以这样做（注意评论）：

#include <iostream>

int main(void) {
  // Declaration and initialization of the required variables
  float cPositive = 0.0f;
  float cNegative = 0.0f;
  int it = 0;

  std::cout << "Enter 10 numbers (floating point assignable): \n";

  // Looping till 10 iterations
  do {
    float temp;
    std::cin >> temp;

    // If the number is greater than zero, i.e. (+ve) then cPositive sums up
    // otherwise, cNegative
    if (temp > 0) cPositive += temp;
    else if (temp <= 0.0f) cNegative -= temp;

  } while (++it < 10); // Increment and comparison together

  // Final results
  std::cout \
    << "Sum of positive: " << cPositive << std::endl
    << "Sum of negative: -" << cNegative << std::endl;

  return 0;
}

简单的测试案例：

Answer 1

Enter 10 numbers (floating point assignable): 
10.5
-1.5
2.2
5.5
-3.8
-99.3
10
4.5
-1.0
0
Sum of positive: 32.7
Sum of negative: -105.6

，如果您想查看平均值，请声明两个变量，即两者最初为零的变量。之后，当发生正数或负数时，只需增加

pos

或neg，然后分别与

pos

或

neg

分别分开。

下面的代码产生以下输出：

cPositive

请阅读评论，因为它们实际上是详细的答案。

cNegative 使用两个变量

$ ./main 
The (sum, avg) of negative integers = (-15, -5)
The (sum, avg) of positive integers = (6, 2)
The (sum, avg) of all numbers = (-9, -1.5)

Answer 2

#include <array>
#include <iostream>

int main()
{
    // For convenience, keep the numbers in an std::array. std::vector is
    // equally convenient.
    std::array<int, 6> integers { 1, -4, 2, -5, 3, -6 };

    // Define variables that store the sums and the counts.
    int positiveSum = 0;
    int positiveCnt = 0;
    int negativeSum = 0;
    int negativeCnt = 0;

    // Iterate over the numbers taking one of them at a time.
    int i = 0;
    while (i < integers.size())
    {
        int number = integers[i];

        // Is the number positive?...
        if (number >= 0)
        {
            // ... it is - add it to the positive sum and increment the count.
            positiveSum += number;
            ++positiveCnt;
        }
        // The number is not positive, so it must be negative...
        else
        {
            // ... add it to the negative sum and increment the count.
            negativeSum += number;
            ++negativeCnt;
        }
       
        // Get ready for the next number.
        ++i;
    }

    // Time to print out the results.
    // Note that before we calculate the average, we have to cast at least one
    // of the terms of the division to floating point type. Otherwise the
    // division will be done with integers where the result is also an integer
    // (e.g. 3 / 2 -> 1).
    // Only affter the casting you will be getting expected answers
    // (e.g. double(3) / 2 -> 1.5).
    std::cout << 
        "The (sum, avg) of negative integers = (" << 
        negativeSum << ", " << 
        double(negativeSum) / negativeCnt << ")" << std::endl;
    
    std::cout << 
        "The (sum, avg) of positive integers = (" << 
        positiveSum << ", " << 
        double(positiveSum) / positiveCnt << ")" << std::endl;

    std::cout << 
        "The (sum, avg) of all numbers = (" << 
        negativeSum + positiveSum << ", " << 
        double(negativeSum + positiveSum) / (negativeCnt + positiveCnt) << ")" << std::endl;
}

。

在循环中尝试一个条件

int negativeVar=0

检测给定数字为负或正值。

然后分别添加所有正值和负值并使平均值。

PositiveVar=0

Answer 3