如何制作“位置”？在我的代码更精确？ [英] How do I make the "position" in my code more precise?

问题描述

我终于为一个赋值完成了这个代码但是有一个问题：对于输出Positions：我希望它输出最小值发生的每个位置。但是，我的代码只输出每个随机生成的数组中的最小位置（大小为20）。

I finally finished this code for an assignment but there is one problem: for the output "Positions:" I want it to output every position where the minimum occurs. However, my code only outputs the last position of the minimum in each randomly generated array (size 20).

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

double random(unsigned int &seed);
unsigned int seed = (unsigned int)time(0);
const int SIZE = 20;

void print_array (int a[]);
void fill_array (int a []);
int find_min (int [], int);

int main ()
{
    int arr[SIZE];
    cout << "Arrays: \n";
    fill_array(arr);
    print_array(arr);
    
    int pos = find_min(arr, SIZE); 
    int minimum = arr[pos];
    
    cout << "Min is: " << minimum << endl;
    cout << "At position: " << pos +1 << endl; 
         
    return 0;
}

double random(unsigned int &seed)
{
    const int MODULUS = 15749;
    const int MULTIPLIER = 69069;
    const int INCREMENT = 1;
    seed = ((MULTIPLIER*seed)+INCREMENT)%MODULUS;
    return double(seed)/MODULUS;
}

void fill_array (int a [])
{
    for (int i = 0; i < SIZE; ++i)
        a[i] = 0 + (10 * (random(seed)));
}

int find_min (int arr[], int n)
{
    int min = arr[0]; 
    int index = 0;
    
    for (int i = 1; i < n; ++i)
        if (arr[i] < min) 
        {
            index = i;
            min = arr[i];
        }
    return index;
}

void print_array (int a[])
{
    for (int i = 0; i <SIZE; ++i)
        cout << setw(3) << a[i];
    cout << endl;
}

我的尝试：



我不明白如何添加更多位置值，因为数组大小为20，无论如何，最小值将至少打印两次。

What I have tried:

I don't understand how to add more position values since the array size is 20 and no matter what, the minimum will be printed at least twice.

推荐答案

编写另一个函数：FindNext。

交给它数组，它的大小，要查找的值，以及开始的索引。

如果找到另一个例子，返回索引。如果不这样做，请返回-1。

然后循环调用并打印最小值，直到找不到它为止。



但这是你的作业，所以我不会给你任何代码！

Write another function: FindNext.
Hand it the array and it's size, the value to look for, and the index to start at.
If you find another example, return the index. If you don't, return -1.
Then loop calling that and printing minima until it can't find any more.

But this is your homework, so I'll give you no code!

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