10966 getting tle in AGS plzz help

[speed_navneet]

t = int(raw_input())
while(t != 0):
	a,d,r = raw_input().split()
	n,mod = raw_input().split()
	a = int(a)
	mod = int(mod)
	d = int(d)
	r = int(r)
	n = int(n)
	if( r == 1):
		print (a+(n/2)*d)%mod
	elif(r != 1):
	        p = (n-1)/2
	        r1 = r-1
		s = (d*r*(r**p - 1))/r1
		res = a*(r**p) + s
		if(n % 2 == 1 and r != 1):
			print res%mod
		else:
			print (res+d)%mod
	t = t-1

getting tle... ???? anything faster could be the soluthion ?? plzz help

[daft_wullie]

@speed.navneet: Did you take into account that raw_input and print are known as slow ?

I use something like

import sys
# ...
cases = sys.stdin.read().split('\n')
# ... processing
sys.stdout.writelines ('\n'.join(results))

as a basic framework for all my python submissions.
cheers
Daft

[speed_navneet]

import sys
t = int(sys.stdin.readline())
while(t != 0):
	a,d,r = sys.stdin.readline().split()
	n,mod = sys.stdin.readline().split()
	a = int(a)
	mod = int(mod)
	d = int(d)
	r = int(r)
	n = int(n)
	if( r == 1):
		sys.stdout.write(str((a+(n/2)*d)%mod)+'\n')
	elif(r != 1):
	        p = (n-1)/2
	        r1 = r-1
		res = (((r**p)*(d*r + a*r1))/r1)-((d*r)/r1)
		if(n % 2 == 1 and r != 1):
			sys.stdout.write(str(res%mod)+'\n')
		else:
			sys.stdout.write(str((res+d)%mod)+'\n')
	t = t-1

i hanged the codes twice but still tle... why time so strict.. help me out plzzzz

[numerix]

i hanged the codes twice but still tle... why time so strict.. help me out plzzzz

Time limit isn't strict - there are no special optimizations necessary to get a Python solution AC.
Your algorithm is not suitable to handle input data with the given constraints.
Just check your code for some testcases near the upper limit of 10⁸ ...

[speed_navneet]

is three any way to escape such calculations????

[numerix]

is three any way to escape such calculations????

Sure - you can see my AC Python code in the submission list of that problem.
10 lines of Python are enough, but the real task is not the implementation, but the development of an efficient algorithm.

[speed_navneet]

@numerix i am not able to see anyones submitted code..

[numerix]

@numerix i am not able to see anyones submitted code..

Of course not the code itself, but the fact that there is an AC submission.

[speed_navneet]

still i tried... no effect else tle....