如何找到随机整数中每个数字的出现次数？

Question

我做了一个简单的随机程序，生成一个 0 到 9 之间的随机数

int random_x;

int main(int argc, char* argv[])
{    
 srand ( time(NULL) );

for (int t=0;t<10;t++)
{ 
    random_x = rand() % 10; //generate 10 random integers between 0 and 10
    cout << "\n" << random_x;
}

下一个阶段是设置一个整数数组来计算每个数字出现了多少。例如，如果随机数是 144566888，那么输出将是

1 appears 1 time 
4 appears 2 times
5 appears 1 time
6 appears 2 times
8 appears 3 times

这是我尝试过的，但它返回“数字出现 0 次”

int count[random_x] = {0};

while ( random_x > 0) {  //loop to return how many times a number appears
    count[ random_x % 10]++;
     random_x =  random_x / 10;
     cout <<"Digit occurs: " << random_x << " times" << endl;
}

Answer 1

这个想法只是保持一个计数数组（最初都是零）并在生成每个数字时更新它们。然后，最后，输出这些计数。

这样的事情会是一个好的开始：

#include <cstdlib>
#include <ctime>
#include <iostream>

int main() {
    srand(time(nullptr));

    // Count accumulators, all originally zero.

    int count[10];
    for (int i = 0; i < 10; ++i) { // or 'int count[10] = {0}' above.
        count[i] = 0;
    }

    // Process random digits, printing and accumulating.

    std::cout << "Generating random digits:";
    for (int i = 0; i < 10; ++i) {
        int digit = rand() % 10;
        std::cout << " " << digit;
        count[digit]++;
    }
    std::cout << "\n";

    // Output the statistics.

    for (int i = 0; i < 10; ++i) {
        if (count[i] == 1) {
            std::cout << "    Digit " << i << " occurs 1 time.\n";
        }
        else if (count[i] > 0) {
            std::cout << "    Digit " << i << " occurs " << count[i] << " times.\n";
        }
    }
}

正如预期的那样，几次运行的输出是：

Generating random digits: 6 1 5 8 9 2 8 0 5 3
    Digit 0 occurs 1 time.
    Digit 1 occurs 1 time.
    Digit 2 occurs 1 time.
    Digit 3 occurs 1 time.
    Digit 5 occurs 2 times.
    Digit 6 occurs 1 time.
    Digit 8 occurs 2 times.
    Digit 9 occurs 1 time.

Generating random digits: 3 4 0 1 3 3 9 9 5 7
    Digit 0 occurs 1 time.
    Digit 1 occurs 1 time.
    Digit 3 occurs 3 times.
    Digit 4 occurs 1 time.
    Digit 5 occurs 1 time.
    Digit 7 occurs 1 time.
    Digit 9 occurs 2 times.

Generating random digits: 7 7 9 7 3 7 4 3 7 4
    Digit 3 occurs 2 times.
    Digit 4 occurs 2 times.
    Digit 7 occurs 5 times.
    Digit 9 occurs 1 time.

Answer 2

您的问题与计算数组中每个唯一位的出现次数有关。

对此有不同的方法，一种是使用unordered_set（存储唯一值）来存储不同的元素并计算唯一元素的总数来获得数组大小，这样我们就可以为每个不同的元素分配一个计数器。

由于您不受时间限制，因此更容易实现两个循环，外部循环遍历所有元素，内部循环首先检查重复项（使用数组检查频率状态）并分别计算每个唯一元素的出现. （然后将计数值传输到频率保持数组，并在每次迭代时相应地打印值）

// your rng-based digits in an array:
int a[9]={1,4,4,5,6,6,8,8,8};

// For checking unique/distinct digits first, and later on to hold the frequency as transferred by count:
int frequency[9]={-1,-1,-1,-1,-1,-1,-1,-1,-1};

    // Our counter variable, but counts will be transferred to frequency[] later on:
    int count;

    for(int i=0; i<9; i++)
    {   // Each distinct element occurs once atleast so initialize to one:
        count = 1;
        for(int j=i+1; j<9; j++)
        {
            // If dupe is found: 
            if(a[i]==a[j])
            {
                count++;

                // Ensuring not to count frequency of same element again:
                frequency[j] = 0;
            }
        }

        // If frequency of current element is not counted:
        if(frequency[i] != 0)
            frequency[i] = count;
    }

    for(int i=0; i<9; i++)
    {
        if(frequency[i] != 0)
            printf("%d appears %d times\n", a[i], frequency[i]);
    }