C How to modify it correctly so that ABCD can be entered

abcd is a character, so comparisons should be enclosed in single quotation marks.

For example. select == 'c'

The correct result is: a

a p=“abcd”;p is a character pointer that gives the address of "abcd" to p, that is, p points to "abcd".", which is pointing to 'a'the address;

b a=“abcd”；a is the address of an array of characters, which cannot be assigned;

c *p=“abcd”;p is a character, and a string cannot be assigned to it, if it is *p='a', is correct.

1. The content of a variable can only be a constant;

2. Strings are not constants, characters are constants;

3. The first address of the string is also a constant, in fact, the first address of the string is a pointer.

Baked: 1Right-click on the input method.

2.Select the first item, Set Properties

3.Select the Common Settings page.

4.In the top input style, remove the check mark in front of "word by space" and "letter word selection" and confirm.

The setting is complete, give points hehe.

You're also a wonder, and people think the letters are convenient.

There is a custom phrase in sogou.

You can try it yourself.

Finally, after c, 100 + 4 = 104

Then there is a, 104-5 = 99

Then there is d, 99 3 = 33

First pass b, 33*2=66

So it's 66

Solution: First of all, it is easy to prove that the four white triangles are congruent, and according to the symmetry of the square, it can be found that the intersection of the two shaded parts forms a square. In this way, the problem is much simpler, let De be Hc and AF at M and N points, and GB be Hc and AF respectively at P and Q

According to the symmetry mentioned above, we know that the MNQP in the middle is a square.

Let ae be x and an be y, since an is perpendicular to en, it is obvious that ane is a right triangle, and according to the congruence, the other 3 white triangles are also right triangles.

Since the 4 triangles are congruent, the area of the triangle ane is 3 4

and en=root number (x 2-y 2).

Thus the area equation by ane an*en 2 = 3 4

Get y*root number(x 2-y 2) 2=3 4 ·· 1）

Also, note: the triangle ane is similar to the triangle aqe, so.

ae/ab=en/bq

And we know that ae=x, ab=2, bq=an=y

Thus, we get x 2 = root number (x 2-y 2) y · · 2）

According to equations (1) and (2), we can reduce y to get x=3 y 2 and bring it into (2).

Getting the equation about x and simplifying it gives us 4x 3-3x 2-12=0

This is a cubic equation, you can use the formula to find the root 3 times, or you can first determine the approximate position of the root, and then use the 2 division to iterate to find the approximate solution, I used maple to calculate, there is only one real root x=, and the other two are imaginary roots, rounded off.

So the final result is x = approximate solution).

In other words, after a second, the area of the blank part in the figure is 3 square centimeters.

Solution: The perpendicular line of the F to do ch is k, and the departure is set to x seconds, obviously, the area of the square = fk 2

and the area of the parallelogram AHCF = CF * CD = CH * FK = > FK = CF * CD CH

Apparently cd=2 cf=x ch=sqrt(cd 2 + dh 2) = sqrt(4 + (2-x) 2).

1 = area of the shaded part of the figure = 2 * area of the parallelogram AHCF - area of the small square.

2cf*cd-fk^2=2*x*2-(x*2/sqrt(4+(2-x)^2))^2

4x-4x^2/(4+(2-x)^2)

4x^3-21x^2+36x-8=0

Iterative calculations are about x.

The above calculation is based on a velocity of 1 cm per second, if not, it is sufficient to divide the result by the velocity.

Well, long waiting.

In your diagram, we know that the parallelogram dgeb area = 4x while in my diagram, the rectangular dega area = 4x

So we think of it as a parallelogram dgeb that translates to get a rectangular dega and a parallelogram chif that becomes a rectangular habf

The area of the square in the middle is equal: because the distance between the parallel lines is equal everywhere, it can be seen as a square that has been rotated and translated to the position of my figure, and then I tried it out:

cd=2, so cg=2-x

Blank area = blank square area = (2-x) = 3x -4x + 4 = 3

x²-4x+1=0

Using the root finding formula, we get:

x = 2 plus or minus (root number 3).

Because x<2

So =2 - root number 3

As for why you calculate the divisor, I have calculated that 2-root number 3 is indeed the divisor for the top floor, =2cf*cd-fk 2=2*x*2-(x*2 sqrt(4+(2-x) 2)) 2 This step should be problematic, because I did the same in the first place, and it was because my teacher said that he used translation.