Of new consult about Python's problem!!!-CodePudding

Crossed the algorithm is why?
Break the specific role here is what?

CodePudding user response:

Friendship remind the landlord to send the wrong place
The role of the break is generally out of the current loop only jump a layer of oh you have layer 2 cycles
See you this code should be out for j in range (2 k + 1) the function of the cycle
As for the marking place you can take all the python I'm not familiar with the==can only look at the understanding

CodePudding user response:

reference 1st floor Li Mu general response:

friendship remind the landlord to send the wrong place
The role of the break is generally out of the current loop only jump a layer of oh you have layer 2 cycles
See you this code should be out for j in range (2 k + 1) the function of the cycle
As for the marking place you can take all the python I'm not familiar with==can only see understand

Ah, thank you for reminding me, I like to send the wrong figure [his face] underlined part k to the if these three lines

CodePudding user response:

A prime number is prime. In addition to it can expressed as the product of its own and 1, can not be expressed as the product of any two integers
Math. SQRT (11) returns the square root of 11="3.316...//square root function is to reduce the cycle of n * m square root of the maximum value is n
Math. Ceil (Math. SQRT (11)) take up the whole="4
Such as prime number 11 assumptions don't know if he is a prime we can do according to the definition judgment
It in addition to outside, expressed as the product of its own and 1 cannot be expressed as the product of any two integers
So in remove 1 from 2 start cycle that is the range (2 K + 1)
the cause ofI % j 11 divisible (2-5) cycle, have been divisible data exist, just skip, i++ directly into the I=12 loop, no output 11 11 is obviously not 11 is a prime number is divisible so i++ until I=30 output and cycle

CodePudding user response:

refer to fifth floor Li Mu general response:

prime number is prime. It is in addition to outside, expressed as the product of its own and 1 cannot be expressed as the product of any two integers
Math. SQRT (11) returns the square root of 11="3.316...//square root function is to reduce the cycle of n * m square root of the maximum value is n
Math. Ceil (Math. SQRT (11)) take up the whole="4
Such as prime number 11 assumptions don't know if he is a prime we can do according to the definition judgment
It in addition to outside, expressed as the product of its own and 1 cannot be expressed as the product of any two integers
So in remove 1 from 2 start cycle that is the range (2 K + 1)
the cause ofI % j 11 divisible (2-5) cycle, have been divisible data exist, just skip, i++ directly into the I=12 loop, output without obvious 11 November 11 is a prime number is not divisible so be output and loop i++ until I=30

Oh oh, that is, as long as the number of (10, 30) aliquot (2 k + 1) in any number of cycle to the next, not out of the result?

CodePudding user response:

refer to 6th floor I very fierce!!!!!! Reply:

Quote: refer to fifth floor Li Mu general response:
prime number is prime. It in addition to outside, expressed as the product of its own and 1 cannot be expressed as the product of any two integers
Math. SQRT (11) returns the square root of 11="3.316...//square root function is to reduce the cycle of n * m square root of the maximum value is n
Math. Ceil (Math. SQRT (11)) take up the whole="4
Such as prime number 11 assumptions don't know if he is a prime we can do according to the definition judgment
It in addition to outside, expressed as the product of its own and 1 cannot be expressed as the product of any two integers
So in remove 1 from 2 start cycle that is the range (2 K + 1)
the cause ofI % j 11 divisible (2-5) cycle, have been divisible data exist, just skip, i++ directly into the I=12 loop, output without obvious 11 November 11 is a prime number is not divisible so be output and loop i++ until I=30

Oh oh, that is, as long as the number of (10, 30) aliquot (2 k + 1) in any number of cycle to the next, not out of the result?

10 to 30 and circulation, the value of k is changing... Don't know you understand not...

CodePudding user response:

refer to 7th floor Li Mu general response:

Quote: refer to the sixth floor I very fierce!!!!!! Reply:

Quote: refer to fifth floor Li Mu general response:
prime number is prime. It in addition to outside, expressed as the product of its own and 1 cannot be expressed as the product of any two integers
Math. SQRT (11) returns the square root of 11="3.316...//square root function is to reduce the cycle of n * m square root of the maximum value is n
Math. Ceil (Math. SQRT (11)) take up the whole="4
Such as prime number 11 assumptions don't know if he is a prime we can do according to the definition judgment
It in addition to outside, expressed as the product of its own and 1 cannot be expressed as the product of any two integers
So in remove 1 from 2 start cycle that is the range (2 K + 1)
the cause ofI % j 11 divisible (2-5) cycle, have been divisible data exist, just skip, i++ directly into the I=12 loop, output without obvious 11 November 11 is a prime number is not divisible so be output and loop i++ until I=30

Oh oh, that is, as long as the number of (10, 30) aliquot (2 k + 1) in any number of cycle to the next, not out of the result?

10 to 30 and circulation, the value of k is changing... Don't know you understand not...

Uh huh, cycle I know, thank you

CodePudding user response:

refer to the eighth floor I very fierce!!!!!! Response:

Quote: refer to 7th floor Li Mu general response:

Quote: refer to the sixth floor I very fierce!!!!!! Reply:

Quote: refer to fifth floor Li Mu general response:
prime number is prime. It in addition to outside, expressed as the product of its own and 1 cannot be expressed as the product of any two integers
Math. SQRT (11) returns the square root of 11="3.316...//square root function is to reduce the cycle of n * m square root of the maximum value is n
Math. Ceil (Math. SQRT (11)) take up the whole="4
Such as prime number 11 assumptions don't know if he is a prime we can do according to the definition judgment
It in addition to outside, expressed as the product of its own and 1 cannot be expressed as the product of any two integers
So in remove 1 from 2 start cycle that is the range (2 K + 1)
the cause ofI % j 11 divisible (2-5) cycle, have been divisible data exist, just skip, i++ directly into the I=12 loop, output without obvious 11 November 11 is a prime number is not divisible so be output and loop i++ until I=30

Oh oh, that is, as long as the number of (10, 30) aliquot (2 k + 1) in any number of cycle to the next, not out of the result?

10 to 30 and circulation, the value of k is changing... Don't know you understand not...

Uh-huh, cycle I know, thank you!

feel you still didn't understand...

CodePudding user response:

Feel you should understand is why most square root parameters
Math. Ceil (Math. SQRT (I))

CodePudding user response:

The prime, have an aliquot judgment, that is not a prime number, the back of the loop there is no need to go on