Finding a C language master!!!! Note the unit oh 20

I feel very simple...

#include

#include

void main()

int l1=100,l2=300,e=30,w1=10;

double q1,q2,w2,a2,v3,a3;//q1=ψ1,q2=θ2

printf("Enter the value of the variable q1:");

scanf("%f",&q1);

q2=asin((l1*sin(q1)-e)/l2);

w2=l1*w1*cos(q1)/(l2*cos(q2));

a2=(l2*w2*w2*sin(q2)-l1*w1*w1*sin(q1))/(l2*cos(q2));

v3=-l1*w1*sin(q1)+l2*w2*sin(q2);

a3=-l1*w1*w1*cos(q1)+l2*w2*w2*cos(q2)+l2*a2*sin(q2);

Why do you have to use C? It should be easier to use MATLAB.

At the beginning of the nth hour, the minute hand is at 12 o'clock and the hour hand is 30n degrees from 12 o'clock, the speed of the minute hand is 6 degrees per minute, and the speed of the hour hand is degrees per minute, so if it coincides at minute m, then it coincides at 6m = 30n + and m = 30n minutes.

The integer part is the number of minutes, and the decimal part is multiplied by 60 to be the number of seconds.

#include

int main()

From the identity of the topic: a=3, b=-1, m=4

f（x）=(x+4)/(x+1)

f（2s+1/s)

6s+1)/(3s+1)

2t-1/t

t=-3s-1

The existence function t=v(s)=-3s-1 (s is greater than 0), satisfying f(2s+1 s)=2t-1 t

The solution is obtained from the identity of the problem: a=3, b=-1, m=4 because f(x)=(x+4) (x+1).

So f(2s+1 s).

6s+1)/(3s+1)

2t-1/t

So t=-3s-1

So there is a function t= (s)=-3s-1 (s is greater than 0), satisfying f(2s+1 s)=2t-1 t

This question is simply wrong. There is no answer.

Solution: f(x)=x+m x+1.

f(2t-1/t）=(2t-1/t)+m/(2t-1/t）+1

It can be obtained from s=(t)=at+b.

2s+1/s=2（at+b）+1/(at+b)=2at+2b+1/(at+b)

and then by f(2t-1 t)=2s+1 s

Get (2t-1 t) + m (2t-1 t) + 1 = 2at+2b+1 (at+b).

Corresponding by the equation, i.e., the coefficients of the same exponent on both sides of the equation are the same, 2t-1 t)+m (2t-1 t)+1=2at+2b+1 (at+b).

When (2t-1 t)*(at+b) is not equal to 0, the equation is multiplied by (2t-1 t)*(at+b) on both sides of the equation.

From the same coefficients of the same exponent, a system of 6 equations can be obtained.

The system of equations is as follows.

1）t3：4a=4aa

2）t2：4b=4ab+4ab

3）t：am-2a-2a=4bb-2aa+2

4) Constant: bm-2b-2b=-2ab-2ab

5）t-1：a=-2b-1

6）t-2：b=0

From 5 of them, a=1, b=0, m=4 can be obtained. Solve as in equation (5) and get: -1=1;

Therefore, there is no solution to the system of equations.

Therefore, this question is simply wrong. There is no answer.

P.S. Prove that a=3, b=-1, m=4 are not the answers.

Let t=1; x=2t-1/t=1;

s=a+b;

f(x)=2+m=2a+2b+1/(a+b);

Obviously, a=3, b=-1, m=4, does not conform to this equation.

If m=4, then: 6=2(a+b)+1(a+b).

Let a+b=y, i.e., y+1 y=6; y2-6y+1=0;

Obviously, a+b is not equal to 2;

Solution: f(x)=x+m x+1.

f(2t-1 t)=(2t-1 t)+m(2t-1 t)+1 is obtained by s= (t)=at+b.

2s+1/s=2（at+b）+1/(at+b)=2at+2b+1/(at+b)

and then by f(2t-1 t)=2s+1 s

Get (2t-1 t) + m (2t-1 t) + 1 = 2at+2b+1 (at+b).

Corresponding to equality from the equation yields m=4

Hello landlord.

The case under VC or GCC is calculated according to the structure of the tree

Calculate (i++)i), the assembly language will first execute ++i, i becomes 4, and then since the return value of i++ is the same as i, it directly performs 4*4 to get 16.

Then calculate (i--)i), the assembly language will first execute --i, i becomes 3, and then because the return value of i-- is the same as i, it directly performs 3*3 9

Then 16+9 equals 25.

Finally, there is a ++ and a --for the final modification of i, and i becomes 3.

This is the case under VC or GCC. Since the optimization performed by the compiler in the process of generating the sink code from the source to the generation of the sink code is different, the calculation results will be different.

Summary: When you are really programming, you must not write this, and do not place multiple auto-increment or auto-decrement operations of the same variable in the same expression.

a=(i++)i)+(i--)i);This statement is ambiguous, and the results are different in different compilation systems. There's no need to dig deeper.

Asking for popular science I can't count 25 and 18

Definition operators have different precedences. It's about the compiler.