您好，我需要有关C ++程序的帮助。 [英] Hello, I need help with a C++ program.

问题描述

我需要编写一个程序，其输出应该是这样的，数组的每个第一个元素都是1，每个第二个元素是0.我需要检查数组中的元素是否为奇数然后打印出1并且如果是元素是奇数然后打印出来0.我需要在条件中检查这个条件。就像有限状态机一样。我没有这样做。任何帮助将不胜感激。



我尝试过：

I need to write a program, Whose output should be like, every first element of an array is 1 and every second element is 0. I need to check in the array if the element is odd then print out 1 and if the element is odd then print out 0. I need to check this condition in if condition.Just like Finite state machine. I am failing in doing so. Any help would be appreciated.

What I have tried:

using namespace std;

int main(){
	
	cout<<"Enter the size of array"<<"\n";
	
	int n,i;
	cin>>n;
	int array[n];
	for( i=0; i<n;i++){
	if(array[i] % 2!=0 && array[i] % 2==0){
		
	}
	cout<<"1,0";
			
}

}

推荐答案

另请参阅我的评论。假设您需要为奇数打印1而对偶数打印0，如果，则甚至不需要。



这是因为

1％2结果1

2％2结果0

3％2结果为1

4％2结果0

...依此类推。咨询例如维基有关模数运算符的详细信息。





版本I 检查数组值

现在我不熟悉 cout 但这样的事情应该可以胜任：



See also my comments. Assuming you Need to print "1" for odd numbers and "0" for even number you don't even Need an if.

This because
1 % 2 results in 1
2 % 2 results in 0
3 % 2 results in 1
4 % 2 results in 0
... and so on. Consult e.g. wiki for detailed Information about the Modulo Operator.


Version I Check Array values
Now I'm not familar with cout but something like this should do the job:

for(i= 0; i < n; i++)
{
   cout << array[i] % 2;
}

版本II 检查索引

Version II Check index

for(i= 0; i < n; i++)
{
   cout << i % 2;
}

版本III 检查索引'

Version III Check index'

for(i= 0; i < n; i++)
{
   cout << (i + 1) % 2;
}

注意：请记住理查德MacCutchan - 专业档案 [ ^ ]：你的主要问题是你永远不会在数组中输入任何值。

Note: Keep in mind the comment of Richard MacCutchan - Professional Profile[^] : "Your main problem is that you never enter any values into the array."

如果要输入数组大小，则需要分配动态内存。

When you want to enter the array size you need to allocate dynamic memory.

int *array = new int[n];
//work it 
delete array;

你需要一个循环来填充数组。



提示：在操作中使用大括号以提高清晰度并避免错误。

And you need a loop to fill the array.

Tip: use braces in your operations for clarity and avoiding bugs.

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