Question

我正在尝试编写程序来查找数字的频率而不要求用户输入。要清楚，我正在寻找一个用户不输入他需要计算的值的想法频率，编译器应默认告诉列表中每个数字的频率而不是用户给出的数字。下面提到的是我的代码。

 #include <stdio.h>
#include<conio.h>

float frequency (int theArray [ ], int number, int x)
 {
int count = 0;
float mount=0.0;
float tot = (float)number;
int u;
int k;
float q,h;

   //   printf("%d",number);

 for (u = 0; u < number; u++)
 {
    if ( theArray[u]==x)
        count++;
 }
  printf ("\nThe frequency of %d in your array is %d ",x,count);

for(k = 0; k< number; k++)
{
    if(theArray[k]==x)

           //count(theArray[k])         
    {       mount++;

    }
  }
  if(mount>1)
    {


    q = mount/tot;


    return q;

    }
   else
  {


   h = mount/tot;

    return h;
    }
  } 


 void main()
   {
   FILE*file = fopen("num.txt","r");
   int integers[100];
   int theArray[100];
   int i=0;
   float e;
   int num;
   int x,k;
   while(fscanf(file,"%d",&num)>0)
   {
    integers[i]=num;
    printf("\n%d",integers[i]);
    i++;
    }

 printf ("\n OK, Thanks! Now What Number Do You Want To Search For Frequency In Your Array? ");
 scanf(" %d", &x);/*Stores Number To Search For Frequency*/
 e =  frequency(integers,i,x); 
printf("\n probability of %d is %f",x,e);
getch();
fclose(file);
     }

所以我现在的输出是： 1）数字列表 2）出现次数的频率 3）我的号码的可能性相反，我正在寻找能够给出我的号码的概率和频率而没有任何内容的东西用户输入。即我正在寻找像

这样的东西

输出：这个数字的频率是----并且它的概率是---------

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Answer 1

这个循环可以解决你的问题

for(i = 0; i < number; i++)
{
    if(array[i] == -999) // checking if the number has been counted before or not
        continue;
    temp = array[i]; // take each number from the array one by one
    for(j = 0; j < number; j++) // iterate over the array to check the number of occurences of temp
    {
        if(array[j] == temp) 
        {
            array[j] = -999; // putting -999 at places where the number has already been counted
            count++;
        }
    }
    printf("frequency of %d is %d, and probability is %f", temp, count, count/number);
    count = 0;
}

如何在不要求用户输入输入的情况下查找数字的频率？

1 个答案: