C language problem: finding the least common multiple and the greatest common divisor of two numbers

C language problem: finding the least common multiple and the greatest common divisor of two numbers

Main() {int x, y, a, m, N, t; printf ("please input two num x, y, n"); scanf (% D,% d ", & X, & Y); for (a = x0; a --) if (x% a = = 0 & & Y% a = = 0) {printf (" the greatest common divisor of X, y is% D, n ", a); break;} M = x / A; n = Y / A; t = a * m * n; printf (" the least common multiple of X, y is
Write two functions to calculate the greatest common divisor and the least common multiple of two integers respectively, call these two functions with the main function and output the results. The two integers are input from the keyboard in the main function
#includeint maxY(int m,int n){int i;for(i=n;i>0;i--) if(m%i==0&&n%i==0)break;return(i);}int minB(int m,int n){int j;for(j=n;j>0;j++) if(j%m==0&&j%n==0)break;return(j);}int main(){int m,n,i,j;scanf...
It is known that M belongs to R, and the complex number Z = m (m-2) / M-1 + (m square + 2m-3) I, when m is what
(1) Z belongs to R
(2) Z is a pure imaginary number
(3) Z is the corresponding point in the second quadrant of the complex plane
(4) The point corresponding to Z is on the line x + y + 3 = 0
(1) M2 + 2m-3 = 0
(m-1) (M + 3) = 0, M-1 = 0, meaningless
m=-3
（2）m（m-2）=0
M = 0 or 2
(3) The corresponding point is in the second quadrant of the complex plane
Then M (m-2) / M-1 ＜ 0, (M2 + 2m-3) ＞ 0
So - 3 < m < 0
(4) M (m-2) / M-1 + (M2 + 2m-3) + 3 = 0
Then M (m-2) / M-1 = M2 + 2m
（m-2）/m-1=m+2
The equation is: M = - 1 + radical 5, or - 1 - radical 5
1)m^2+2m-3=0,(m+3)*(m-1)=0,∴m=1&-3
2) If M (m-2) = 0, M = 1 or 2, and m ^ 2 + 2m-3 is not equal to 0, then 0 and 2 are OK
3) M (m-2) / (m-1) < 0, equivalent to m (m-1) (m-2) < 0, m < 0 or 1 < m < 2, and m ^ 2 + 2m-3 > 0, ■ - 3 < m < 1
So - 3 < m < 0
4) M (m-2) / M-1 + (m ^ 2 + 2m-3) = - 3
1)m^2+2m-3=0,(m+3)*(m-1)=0,∴m=1&-3
2) If M (m-2) = 0, M = 1 or 2, and m ^ 2 + 2m-3 is not equal to 0, then 0 and 2 are OK
3) M (m-2) / (m-1) < 0, equivalent to m (m-1) (m-2) < 0, m < 0 or 1 < m < 2, and m ^ 2 + 2m-3 > 0, ■ - 3 < m < 1
So - 3 < m < 0
4) M (m-2) / M-1 + (m ^ 2 + 2m-3) = - 3, sorting out the equation: m (m * m + 2m-4) = 0, M = 0, or m ^ 2 + 2m-4 = 0. M = - 1 + radical 5, or - 1 - radical 5
In the complex plane, the point corresponding to the complex number − 2 + 3i3 − 4I lies in the second plane______ Quadrant
By − 2 + 3i3 − 4I = (− 2 + 3I) (3 + 4I) (3 + 4I) = − 18 + I25 = − 1825 + 125I, the point corresponding to the complex number is (− 1825125), which is the point of the second quadrant. Therefore, the answer is two
English plural of monkey family English plural of fish
Monkey plural: Monkeys
The plural of family
The plural of fish
Basic operation of complex number
Please help me: how to calculate the nth power of (COSA + I * Sina)
This uses Euler's formula
The IA degree of cosa + I * Sina = e ^ (IA) e, e is the base of natural logarithm, e = 2.71828
The nth power of (COSA + I * Sina) is e ^ (INA)
That is cos (NA) + I * sin (NA)
Demover theorem
(cosA+isinA)^n=cos(nA)+isin(nA)
This uses Euler's formula
The IA degree of cosa + I * Sina = e ^ (IA) e, e is the base of natural logarithm, e = 2.71828
The nth power of (COSA + I * Sina) is e ^ (INA)
That is cos (NA) + I * sin (NA)
Or demover theorem
(cosA+isinA)^n=cos(nA)+isin(nA)
When the value of real number m is taken, the point (1) representing complex z = (m2-8m + 15) + (m2-5m-14) I in the complex plane is located in the fourth quadrant? (2) In the first and third quadrants? (3) On the line y = x?
(1) According to the meaning of the title, if M2 − 8m + 15 ＞ 0m2 − 5m − 14 ＜ 0, the solution can get (m − 3) (m − 5) ＞ 0 (m − 7) (M + 2) ＜ 0, that is & nbsp; m ＞ 5 & nbsp; & nbsp; or m ＜ 3 − 2 ＜ m ＜ 7, so - 2 ＜ m ＜ 3, or 5 ＜ m ＜ 7. (2) from (m2-8m + 15) (m2-5m-14) ＞ 0, (M-3) (m-5) (M-7) (M + 2) ＞ 0, so & nbsp; & nbsp; M ＜ - 2, or 3 ＜ m ＜ 5, or m ＞ 7. (3) from m2-8m + 15 = m2-5m-14, we get - 3M = - 29, M = 293
The plural of English words
Frenchmen ['frentʃmən]
Basic translation
n. French (plural of Frenchman)
Network interpretation
French: French
French French man French men parleyvoo
Fifty million French
Let Z satisfy 1 − Z1 + Z = I, then | 1 + Z | = ()
A. 0B. 1C. 2D. 2
Because 1 − Z1 + Z = I, so 1-z = I + Zi, so z = 1 − I1 + I (1 − I) (1 − I) (1 + I) (1 − I) = 2 I2 = − I, then | 1 + Z | = | 1 − I | = 2, so C is selected
Let there be a complex Z satisfying the following conditions: (1) the corresponding point of complex Z in the complex plane is in the second quadrant; (2) Z · Z + 2iz = 8 + AI (a ∈ R)
Let z = m + Ni (m ＜ 0, n ＞ 0) from (1), then from (2), | Z | 2 + 2I (M + Ni) = 8 + AI, that is M2 + n2-2n + 2mi = 8 + AI, ｜ M2 + N2 − 2n = 8 & nbsp; a = 2m & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ② From (1), we get that: M2 + (n-1) 2 = 9, the point Z corresponding to the complex Z is the part of the circle M2 + (n-1) 2 = 9 in the second quadrant, then - 6 ≤ 2m < 0