Write program, seek 1 + 2! + 3! + +The value of 20 Try to write a program, seek 1 + 2! + 3! + +The value of 20! (n! = 1 * 2 * 3 *...) *n） Don't use complicated sentences that I don't understand, You are so talented on the first floor and the second floor. I can't understand it. It's not the answer I want. It's too intelligent. I haven't learned such a high IQ yet~ The structure on the third floor is what I want. It's not so easy to use the nesting of circular sentences~ Would you please make it clear on the fourth floor? I can't really see it~ Mourning

Write program, seek 1 + 2! + 3! + +The value of 20 Try to write a program, seek 1 + 2! + 3! + +The value of 20! (n! = 1 * 2 * 3 *...) *n） Don't use complicated sentences that I don't understand, You are so talented on the first floor and the second floor. I can't understand it. It's not the answer I want. It's too intelligent. I haven't learned such a high IQ yet~ The structure on the third floor is what I want. It's not so easy to use the nesting of circular sentences~ Would you please make it clear on the fourth floor? I can't really see it~ Mourning

1. Write out the output results t ← 2 I ← 5 while t ≤ 2007, t ← T + 1 I ← I + 3 end if print I
2. S ← 0 read n for I from 1 to n Step3 s ← S + I end for print s if the value of input variable is 3, then the value of output variable s ------ if the value of output variable s is 22, then the value of variable s is------
3. Design an algorithm to find the arithmetic mean of 15 numbers and write pseudo code
4. Write pseudo code, find 1 + 1 / (1 * 2) + 1 / (1 * 2 * 3) + 1 / (1 * 2 * 3 * 4). + 1 / (n!) and draw the flow chart
5. Write the algorithm to find the smallest positive integer n satisfying 2 + 5 + 8 + 11 +... + n > 2009

I = 60232. First question: S = 1. Second question: n = 10 or n = 11 or n = 123. S ← 0For I from 1 to 15 step 3read a (I) s = S + a (I) end fors = s / 15print S4. Read ns ← 0t ← 1for I from 1 to t = t * is = S + 1 / ten forprint S5

It is known that the lengths of three sides of a triangle are all integers, and the maximum length is 11. Find the number of triangles satisfying the condition

Because the lengths of three sides of a triangle are all integers and the maximum side length is 11, a limit is a = b = 11, so it is listed as follows: a = 11 b = 11 A = 11 b = 10A = 11 b = 9A = 11 b = 8A = 11 b = 7a = 11 b = 6A = 11 b = 5A = 11 b = 4A = 11 b = 3A = 11 b = 2A = 11 b = 1A = 10 b = 10

If the complement of an acute angle is 40 ° more than twice the remainder of the angle, find the degree of the acute angle

40 degrees

The remainder angle of an acute angle is 1 / 6 of the complement angle of the acute angle. Find an angle 30 ° larger than the degree of the acute angle

Let this angle be X
（90-x)=1/6 (180-x)
x=72
x+30=72+30=102

If the ratio of an acute angle to its residual angle is 5:4, what is the degree of the complement of the acute angle______ ．

Let the acute angle be α, then its remainder angle is 90 ° - α. From the meaning of the question, α 90 ° - α = 54, the solution is: α = 50 °, then the complement angle of α is: 180 ° - α = 130 °. So the answer is: 130 °

The complementary angles of ∠α and ∠β are each other, and the residual angle of ∠α is one seventh of ∠β. The degree of ∠α and ∠β can be calculated

Solving a quadratic equation of two variables

The horizontal degree is equal to () a50% B200% C100% of the right degree

Option B: 200%
 
180÷90×100%=200%
 

1 corner=______ Flat angle=______ Right angle

Because 1 circle angle = 360 °, 1 horizontal angle = 180 ° and 1 right angle = 90 °, 1 circle angle = 2 horizontal angle = 4 right angle

1 circumference = 1 horizontal angle = 1 right angle

1 circumference = 2 horizontal angles = 4 right angles