Enter two positive integers m and N to find their greatest common divisor and least common multiple

Enter two positive integers m and N to find their greatest common divisor and least common multiple

#Include # include int main (void) {int m, N, R; int s; printf ("input two numbers):"; scanf (% d% d ", & M, & n); s = m * n; while (n! = 0) {r = M% n; m = n; n = R;} printf (" GCD% D, n ", m); printf (" LCM% D, n ", S / M); / / system (" paus
C language program problem to find the greatest common divisor and the least common multiple of two integers
The greatest use of division by rotation is to find the greatest common divisor of two numbers. Use (a, b) to express the greatest common divisor of a and B. theorem: A, B, C are positive integers. If a is divided by B, then (a, b) = (B, c). Example: find the greatest common divisor of 15750 and 27216. ∵ 27216 = 15750 × 1 + 11466 ∧ (15
How to solve the equations Z ^ 4 + a = 0 (a > 0) and e ^ (Z + 1) + 2 = 0 in complex field?
We're going to be the z = a ^ (1 / 4) e ^ (I π + 2ik π) / 4, k = 0,1,2,3, and we're going to be the ones that we want to do all of the following (1 / 4) e ^ (1 / 4) e ^ (1 / 4) e ^ (1 / 4) e ^ (1 / 4) e ^ (1 / 4) e ^ (I 3 π / 4) let z = a (1 / 4) e ^ (1 / 4) e ^ (I (I (I π + 2) e ^ (1 / 4) e ^ (I 7 π / 7 π / 4) e ^ (I (7 π / 4) e ^ (7 π / 4) e ^ (7 (I (7 π / 4) e ^ (7) e ^ (7) e \\\\\π + I
Given the complex number Z = 3-4i / M + 2I, the value of real number m is ()
A 8 / 3 B 3 / 2 C negative 8 / 3 D negative 3 / 2
Choose D
From Z1 = m + 2I, Z2 = 3-4i,
Then Z &; / Z &; = m + 2I / 3-4i = (M + 2I) (3 + 4I) / 25 = 3m-8 / 25 + 4m + 6 / 25, I is a real number, and 4m + 6 = 0, so the real number m is - 3 / 2
Z should be a real number, otherwise there is no solution;
If the numerator and denominator are multiplied by 3 + 4I, we can get the number: [3m-8 + (4m + 6) I] / 25. If the imaginary part is equal to 0, we can get m = - 3 / 2, choose D
In the complex range, how many roots of the equation Z ^ 2 + | Z | = 0 (please solve the equation)
Let z = a + IB be substituted into: A ^ 2 + 2abi-b ^ 2 + √ (a ^ 2 + B ^ 2) = 0. Compare the real part with the imaginary part, we get: A ^ 2-B ^ 2 + √ (a ^ 2 + B ^ 2) = 0 1) 2Ab = 0 2) so a = 0 or B = 0 when a = 0, we get: - B ^ 2 + | B | = 0, we get: B = 0,1, - 1 when B = 0, we get: A ^ 2 + | a | = 0, we get: a = 0, so the solution of the original equation is
Z ^ 2 = - LZL, LZ ^ 2L = LZL ^ 2 = LZL, LZL = 0 or LZL = 1. For LZL = 0, z = 0, for LZL = 1, we further have Z ^ 2 = - 1, z = I or Z = - I;
On the contrary, if z = 0, Z ^ 2 + | Z | = 0 satisfies the condition; Z = I or - I, Z ^ 2 + | Z | = 0 also satisfies the condition
To sum up, z = 0, I, - I
Z ^ 2 + | Z | = 0
Let z = x + iy be substituted into the original equation
X ^ 2-Y ^ 2 +2 xyi +√（X ^ 2 + Y ^ 2 ）= 0
So 2XY = 0, x ^ 2-y ^ 2 + √ (x ^ 2 + y ^ 2) = 0
X = 0, y ^ 2 + | y | = 0, we get: | y | = 0 or 1, that is, y = 0, 1, - 1
Expand for y
Z ^ 2 + | Z | = 0
Let z = x + iy be substituted into the original equation
X ^ 2-Y ^ 2 +2 xyi +√（X ^ 2 + Y ^ 2 ）= 0
So 2XY = 0, x ^ 2-y ^ 2 + √ (x ^ 2 + y ^ 2) = 0
X = 0, y ^ 2 + | y | = 0, we get: | y | = 0 or 1, that is, y = 0, 1, - 1
For y = 0, χ ^ 2 + | x | = 0, we get: | x | = 0, that is: x = 0
So share three solutions: z = 0, me, me. Put it away
A={x|x^2-(4+i)x+k+2i=0,k∈R},B={x||x-1+xi≤√（2）^log2(3-2x)|}
A. B is a nonempty set of real numbers
Find K, a, B
x2-(4+i)x+k+2i=0
x2-4x+k+i(2-x)=0
∴x=2
And ∵ K ∈ R
∴k=4
A={2}
√2^(log2(3-2x))=2^[(log2(3-2x))/2]=2^(log2√(3-2x))=√(3-2x)
|x-1+xi≤√（2）^log2(3-2x)|=|x-1+ix≤√(3-2x)|
People think that the sign of absolute value should be enclosed before and after X-1 + iX, indicating the module of the complex number
So the question is:
√((x-1)2+x2)≤√(3-2x)
Both sides square at the same time
2x2-2x+1≤3-2x
x2≤1
B={x|-1≤x≤1}
x^2-(4+i)x+k+2i=0
x^2-4x+k+(2-x)i=0
And X, K ∈ R
∴2-x=0
x^2-4x+k=0
∴x=2
K=4
B what does that mean? The plural number can't be compared in size... Only the mold can...
What curve is the locus of the corresponding point of complex Z on the complex plane satisfying the condition | Z-2 + I | = 3?
Let the vector corresponding to Z in the complex plane be OA = (a, b)
Z1 = 2-I, the corresponding vector in the complex plane is ob = (2, - 1)
Then the formula given in the question is: | z-z1 | = 3
That is: | oa-ob | = 3
That is: | BA | = 3
So, Ba & # 178; = 9
That is: (A-2) &# 178; + (B + 1) &# 178; = 9
Therefore, the locus of the corresponding point of Z in the complex plane is a circle: (X-2) &# 178; + (B + 1) &# 178; = 9
| Z-（0 +）=√（2 +4）= 5
The distance to (0,1) is equal to 5
It's a circle
Selected C
Polar coordinate equation: r = 1 + cos θ?
Transformation to rectangular coordinate system equation
r=1+cosθ
=1+2cos²(θ/2)-1
=2cos²(θ/2)
A = {Z | Z minus 1 plus I | = radical 2}, B = {w | w = 2Z plus 3, Z belongs to a}. Then the complex equation of the curve on the complex plane corresponding to set B is? Sharp
B = {w | w = 5 + 2 radical 2-2i}
What's the difference between ordinary equation, Cartesian equation, parametric equation and polar equation?
This question is not easy to express
My understanding is that the essence is the same, but the expression is different
Different expressions make the geometric meaning of the letters in the equation different
The ordinary equation, that is, the rectangular coordinate equation, is represented by X and Y only
The parameter equation contains the third letter besides X and y, and X and y can be expressed by the expression of this letter
The polar equation does not contain x, y, and is represented by a length P and an angle θ
The transformation method between ordinary equation and polar equation is as follows
Use the following common formulas to transform
X = PCOS θ & nbsp; & nbsp; & nbsp; y = PSIN θ formula: P & # 178; = x & # 178; + Y & # 178; & nbsp; & nbsp; & nbsp; & nbsp; Tan θ = Y / X & nbsp; & nbsp; (x ≠ 0)
For example:
Circle: X & # 178; + Y & # 178; = 4x & nbsp; this is the rectangular coordinate equation (ordinary equation)
After formulation
(X-2) & # 178; + (y-0) & # 178; = 4
X = 2 + 2cost, y = 2sint & nbsp; & nbsp; (using the formula Sin & # 178; a + cos & # 178; a = 1)
Polar equation: ρ = 4cos θ
This is not a matter of distinction, but of condition. According to your list,
1. Ordinary equations have no geometric meaning, and the number and order of unknowns can be defined arbitrarily.
2. The Cartesian coordinate equation includes (but not necessarily includes) two axis elements (generally x, y) of the Cartesian coordinate system with unlimited order.
3. The parametric equation includes (but not necessarily all) all the elements of the number axis in the coordinate system (determined by conditions, such as X, y, Z, u, v...) Besides, it also includes the dependent parameters (determined according to the conditions, such as a, B, C, m, K...) ... unfold
This is not a matter of distinction, but of condition. According to your list,
1. Ordinary equations have no geometric meaning, and the number and order of unknowns can be defined arbitrarily.
2. The Cartesian coordinate equation includes (but not necessarily includes) two axis elements (generally x, y) of the Cartesian coordinate system with unlimited order.
3. The parametric equation includes (but not necessarily all) all the elements of the number axis in the coordinate system (determined by conditions, such as X, y, Z, u, v...) Besides, it also includes the dependent parameters (determined according to the conditions, such as a, B, C, m, K...) The order is unlimited.
4. The polar equation is similar to the rectangular equation except that X and y are replaced by R and o.
For example:
1. The general equation is x ^ 2 + y ^ 3-z / a = 45.
2. The rectangular coordinate equation is y = x / 3 + 2.
3. The parametric equation is x = 2cosa, y = 3sina.
4. The polar equation is in the form of R = 5sin3o. Put it away