How does FORTRAN find the minimum of a set of numbers Use FORTRAN to get a group of numbers, such as a, B, C, D, and then use what statement to get the smallest of the four numbers?

How does FORTRAN find the minimum of a set of numbers Use FORTRAN to get a group of numbers, such as a, B, C, D, and then use what statement to get the smallest of the four numbers?

Is the algorithm upstairs too troublesome?
He should be a novice
In fact, the minimum value of array is very simple
real a(n)
The minimum value of write (*, *) is: ', minval (a)
That's fine
Even if you don't use minval function, it's too troublesome to write like this
A more simple and efficient method is as follows:
mymin = a(1)
do i = 2 ,n
mymin = min( mymin ,a(i) )
end do

FORTRAN language: what do character comet (*) * 60 and character ant * (*) mean?

Character comet (*) * 60 means to define an array of strings, named come. The length of each element is 60, and the number of elements is determined by the arguments. Character ant * (*) means to define a string, named ant, and its length is determined by the arguments_ Thlws Impl...

X+5/7X=4.

X+5/7X=4.8(1+5/7)X=4.8(12/7)X=4.8X=4.8*7/12X=2.8

Given that the equation cos2x + 4sinx-a = 0 has a solution, then the value range of A

The original equation can be transformed into a = cos2x + 4sinx = 1-sin2x + 4sinx = - (sinx-2) 2 + 5 ∵ SiNx ∈ [- 1, 1] ∵ a ∈ [- 4, 4]

How much is the reciprocal of a if the product of a number and its reciprocal plus a is 6 and 1 / 2?

The product of a number and its reciprocal = 1
1+a=13/2
A = 13 / 2-1 = 11 / 2 = 5 and 1 / 2

The equation of a line that is parallel to the line 3x + 4Y = 5 and whose distance is equal to 3 is___ ．

Let the obtained line be 3x + 4Y + M = 0, and the line 3x + 4Y = 5 is 3x + 4y-5 = 0, then d = | m + 5 | 32 + 42 = 3 can be obtained from the distance formula of parallel lines, and the solution is m = 10 or - 20. Then the obtained line is 3x + 4Y + 10 = 0, or 3x + 4y-20 = 0. So the answer is: 3x + 4Y + 10 = 0, or 3x + 4y-20 = 0

Given the function f (x) = 3x-5 / ax ^ 2 + ax + 1. If the domain of F (x) is r, find the range of real number a

F (x) = 3x-5 / ax ^ 2 + ax + 1
The denominator ax ^ 2 + ax + 1 is not zero
When a = 0, denominator = 0 + 0 + 1, which meets the requirements;
When a ≠ 0, x ^ 2 + X + 1 / a ≠ 0, opening upward, extreme value > 0
[(4 * 1 / a) - 1] / 4 > 0, 4 / A-1 > 0, (4-A) / a > 0, equivalent to a (A-4) < 0
0＜a＜4
In conclusion: 0 ≤ a < 4

Parabola y = 2 (X-2) &# 178; the coordinate of intersection a with X axis is, the coordinate of intersection B with y axis is, s △ AOB=

Parabola y = 2 (X-2) &# 178; coordinates of intersection a with X axis are (2,0), coordinates of intersection B with y axis are (0,8), s △ AOB = 8

Given that the axis of symmetry x = - 1 of y = ax ^ 2 + BX + C = 0, the highest point is y = 2x + 4, find the coordinates of the intersection of parabola and straight line

The highest point is the vertex, on the axis of symmetry
x=-1,y=-2+4=2
So the intersection is (- 1,2)

In order to make (x-1) ^ - (x 1) ^ - 2 meaningful, what conditions should the value of X satisfy

(x+1)^(-2)=1/(x+1)^2
Denominator x + 1 ≠ 0
x≠-1