Prime numbers

It means thats the running time for the program, if you compiled his code for 10000 and my code for 10000 my code is like a lot faster

Oh, that's interesting. How did you figure that one out?

Ok you see lala's code:

Code:

 for (prime = 1;prime <=10000; prime++)

Thats about 10000 runs which is n

Now inside that for loop is another for loop!

Code:

for (checker = 1; checker <prime; checker++)

the average of all these runs is about n/2 (5000)

(sorry i was wrong it n*(n/2) not n sqaured)

Now my code:

Code:

for(i=5;i<10000;i+=2)

thats about n/2 (5000) runs for each odd number

Code:

for(i2=0,prime=true;primelist[i2]<=sqrt(i);i2++)

i check the prime list for any factors up to the square root of the number to check

that should be sqrt(n)/2 about the average

so my running time would be: n/2 * sqrt(n)/2 and lalas: n*(n/2)

Originally Posted by zhaoyun333

Ok you see lala's code:

Code:

 for (prime = 1;prime <=10000; prime++)

Thats about 10000 runs which is n

Now inside that for loop is another for loop!

Code:

for (checker = 1; checker <prime; checker++)

the average of all these runs is about n/2 (5000)

(sorry i was wrong it n*(n/2) not n sqaured)

Now my code:

Code:

for(i=5;i<10000;i+=2)

thats about n/2 (5000) runs for each odd number

Code:

for(i2=0,prime=true;primelist[i2]<=sqrt(i);i2++)

i check the prime list for any factors up to the square root of the number to check

that should be sqrt(n)/2 about the average

so my running time would be: n/2 * sqrt(n)/2 and lalas: n*(n/2)

Oh hey zhao is canadian. As am i. Either way i suppose that makes sense. Probably.

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