The bosses can help-CodePudding

CodePudding user response:

# x and y are integer of x and y all possible
# x + y=A
# x * y=B

Def fun (A, B) :
For I in range (A) :
For j in range (B) :
If I + j==A and I * j==B:
Print (I, j)
Fun (8 (9)
Nested for loop can work out the solution

CodePudding user response:

The problem with cycle may be not, because A, B are integers, including positive integer, zero, negative integer
For example: A=7, B=120,
The solution is:
X=15, Y=8
X=8, Y=15

The equation:
X ^ 2 - Ax + B=0

//pseudocode:
The delta=A ^ 2-4 b
If the delta & lt; 0 then 'No solution'
The else
If SQRT (delta) is not an integer AND (A + SQRT (delta) % 2 is not an integer AND (A - SQRT (delta) % 2 is not an integer
Then 'No Solution'
The else
X1=(A + SQRT (A ^ 2-4 b))/2, X2=(A - SQRT (A ^ 2-4 b))/2
If X1==X2 then only one solution: X=X1, Y=X1
Else there are two distinct solutions:
X=X1, X2 and X==Y X2, Y=X1
Note that the root of integer (123.0 to 123)

CodePudding user response:

Both please input A (integer, | A | & lt; 1
1 e7) : -Both please input B (integer, | | B & lt; 12 e14) :
X=3, Y=4
X=4, Y=3

 A=(int) (input (" both please input A (integer, | A | & lt; 1 e7) : ")) 
B=(int) (input (" both please input B (integer, | | B & lt; 1 e14) : ")) 

If (abs (A) & gt;=1 e7 or abs (B) & gt;=1 e14) : 
Print (" Invaild inputs ") 

Def func (A, B) : 
Threshold=2 * * (A) - (A) (B) + 1 
For I in range (threshold) : 
For j in range (threshold) : 
If (I + j==A and I * j==B) : 
Print (" X={}, Y={} ". The format (I, j)) 
Print (" X={}, Y={} ". The format (j, I)) 
Return 
Elif (I + j==and - I * j==B) : 
Print (" X={}, Y={} ". The format (I, j)) 
Print (" X={}, Y={} ". The format (j, I)) 
Return 
Elif (- I - j==A and I * j==B) : 
Print (" X={}, Y={} ". The format (- I, j)) 
Print (" X={}, Y={} ". The format (j, I)) 
Return 
Elif (I - j==and - I * j==B) : 
Print (" X={}, Y={} ". The format (I, j)) 
Print (" X={}, Y={} ". The format (j, I)) 
Return 
The return - 1 

If func (A, B)==1: 
Print (" No solution ")

CodePudding user response:

Judge whether a number is an integer, the built-in function is_integer () is useful:
A=1.2
Anderson s_integer () # False
B=1.0
B.i s_integer (#) True

CodePudding user response:

refer to the second floor rrrr336 response:

this topic using cycle could not, because A and B are integers, including positive integer, zero, negative integer
For example: A=7, B=120,
The solution is:
X=15, Y=8
X=8, Y=15

The equation:
X ^ 2 - Ax + B=0

//pseudocode:
The delta=A ^ 2-4 b
If the delta & lt; 0 then 'No solution'
The else
If SQRT (delta) is not an integer AND (A + SQRT (delta) % 2 is not an integer AND (A - SQRT (delta) % 2 is not an integer
Then 'No Solution'
The else
X1=(A + SQRT (A ^ 2-4 b))/2, X2=(A - SQRT (A ^ 2-4 b))/2
If X1==X2 then only one solution: X=X1, Y=X1
Else there are two distinct solutions:
X=X1, X2 and X==Y X2, Y=X1
Note that the root of integer (123.0 to 123)

Decide there is a little problem, is the is not a "%" :
If SQRT (delta) is not an integer AND (A + SQRT (delta) / 2 is not an integer AND (A - SQRT (delta) / 2 is not an integer