怀疑C ++中的precision（）函数 [英] Doubt regarding precision( ) function in C++

问题描述

给定一个大小为n且多个值的数组，我们需要左右旋转数组。



输入：

第一行由T测试案例组成。每个测试用例的第一行由N和K组成，N表示数组元素的数量，K表示要移位的位数。每个测试用例的第二行由数组元素组成。



输出：

单行输出，打印旋转的数组。



约束条件：

1< = T< = 100

1< = N< = 10 ^ 4
1< = K< = 10 ^ 4



示例：

输入：

1

5 14

1 3 5 7 9

输出：

9 1 3 5 7


我的代码附在这里，虽然我得到了正确的答案但是在期望的输出和预期的输出（关于使用精度函数）方面存在细微差别（.000004），什么我应该在这里修改吗？



我尝试了什么：



我的代码是 -

Given an array of size n and multiple values around which we need to left rotate the array.

Input:
First line consists of T test case. First line of every test case consists of N and K, N denoting number of elements of array and K denoting the number of places to shift. Second line of every test case consists of elements of array.

Output:
Single line output, print the rotated array.

Constraints:
1<=T<=100
1<=N<=10^4
1<=K<=10^4

Example:
Input:
1
5 14
1 3 5 7 9
Output:
9 1 3 5 7

My code is attached here , although I am getting right answer but there is slight difference(.000004) in desired output and expected output (its regarding use of precision function), what should I modify here?

What I have tried:

My code is-

#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
int main()
{

    int t;
    cin>>t;
     
    while(t--)
    {
        int set_size,i;
        cin>>set_size;
        int arr[set_size];
        for(i=0;i<set_size;i++)
        {
            cin>>arr[i];
        }
        
    unsigned int pow_set_size = pow(2, set_size);
	int counter, j;
    float sum_avg=0;
    cout<<fixed<<setprecision(6);
	/*Run from counter 000..0 to 111..1*/
	for(counter = 0; counter < pow_set_size; counter++)
	{
	
           float avg,sum=0;int no=0;
			
	for(j = 0; j < set_size; j++)
	{
		/* Check if jth bit in the counter is set
			If set then pront jth element from set */
			
			
		if(counter & (1<<j))
			{
			      //  cout<<arr[j];
			        sum+=arr[j];
			        no++;
//			        int no=0,sum=0,digit;
//			        float avg;
//			        while(arr[j]!=0)
//			        {
//			          digit=arr[j]%10;
//			          sum=+digit;
//			          arr[j]/=10;
//			          no++;
//			          
//			        }
//			        avg=sum/no;
//			        
//			
//			    overall_sum+=avg;
//			    cout<<"overall_sum="<<overall_sum<<endl;
			}
			
			
			
	}
	
	if(sum!=0)
	{avg=sum/no;
	cout<<"avg= "<<avg;
	    sum_avg+=avg;
	}
	cout<<endl;
	
    }
  // cout<<fixed<<setprecision(6);
   cout<<sum_avg<<endl;
    }
    return 0;
    
}




 //1
//3
//2 3 5

//Its Correct output is:
//23.333333

//And Your Output is:
//23.333334

推荐答案

计算机上的浮点数通常是近似值，它们的精度取决于它们是单长度值还是双倍长度值。请参阅每个计算机科学家应该知道的关于浮点运算的内容 [ ^ ]。

Floting point numbers on a computer are usually approximations, and their precision depends upon whether they are single or double length values. See What Every Computer Scientist Should Know About Floating-Point Arithmetic[^].

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