C 中的随机数和不同数 [英] Random and Different Numbers in C

问题描述

我正在尝试用 C 生成 7 个数字.代码应包含:

I am trying to generate 7 numbers in C. The code should contain:

• 不能以 0 开头.
• 数字必须在 0-9 之间
• 数字必须彼此不同.(例如:不能有两个 5，比如这个:7 5 8 3 2 5 4).

我的代码正在运行.不是从 0 开始的.数字是在 0-9 之间随机生成的.

My code is working. It doesn't start with 0. Numbers are randomly generated between 0-9.

但我无法包含第三件事.当我启动代码时，相同的数字即将到来.你知道我怎样才能产生不同的结果吗?

But I can't manage to include third thing. Same numbers are coming when I start the code. Do you know how can I generate all differently?

#include <stdio.h>
#include <stdlib.h>
#include <conio.h> 
#include <locale.h>
#include <time.h>

int main() {
  srand(time(NULL));
  int numbers[10];
  int i;
  for (i = 0; i < 7; i++) {
    numbers[i] = rand() % 10;
    if (numbers[0] == 0) {
      numbers[0] = 1 + rand() % 9;
    }
    printf(" %d ", numbers[i]);
  }  
  getch();
  return 0;
}

推荐答案

Using permutation (Knuth 洗牌算法)

Using permutation (Knuth shuffle algorithm)

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <time.h>

static int rrand(int range)
{
    return (int)((double)range * (rand() / (RAND_MAX + 1.0)));
}

static void randomize(int arr[], int size)
{
    while (size > 1)
    {
        int item = rrand(size--);
        int temp = arr[size];

        arr[size] = arr[item];
        arr[item] = temp;
    }
}

int main(void)
{
    int arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
    int size = sizeof arr / sizeof *arr;

    srand((unsigned)time(NULL));
    while (arr[0] == 0)
    {
        randomize(arr, size);
    }
    for (int i = 0; i < 7; i++)
    {
        printf("%d\n", arr[i]);
    }
}

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