Approach and Code for o(log n) solution - algorithm

Question

f(N) = 0^0 + 1^1 + 2^2 + 3^3 + 4^4 + ... + N^N.
I want to calculate (f(N) mod M).
These are the constraints.
1 ≤ N ≤ 10^9
1 ≤ M ≤ 10^3
Here is my code
test=int(input())
ans = 0
for cases in range(test):
arr=[int(x) for x in input().split()]
N=arr[0]
mod=arr[1]
#ret=sum([int(y**y) for y in range(N+1)])
#ans=ret
for i in range(1,N+1):
ans = (ans + pow(i,i,mod))%mod
print (ans)
I tried another approach but in vain.
Here is code for that
from functools import reduce
test=int(input())
answer=0
for cases in range(test):
arr=[int(x) for x in input().split()]
N=arr[0]
mod=arr[1]
answer = reduce(lambda K,N: x+pow(N,N), range(1,N+1)) % M
print(answer)

Answer 1

Two suggestions:
Let 0^0 = 1 be what you use. This seems like the best guidance I have for how to handle that.
Compute k^k by multiplying and taking the modulus as you go.
Do an initial pass where all k (not exponents) are changed to k mod M before doing anything else.
While computing (k mod M)^k, if an intermediate result is one you've already visited, you can cut back on the number of iterations to continue by all but up to one additional cycle.
Example: let N = 5 and M = 3. We want to calculate 0^0 + 1^1 + 2^2 + 3^3 + 4^4 + 5^5 (mod 3).
First, we apply suggestion 3. Now we want to calculate 0^0 + 1^1 + 2^2 + 0^3 + 1^4 + 2^5 (mod 3).
Next, we begin evaluating and use suggestion 1 immediately to get 1 + 1 + 2^2 + 0^3 + 1^4 + 2^5 (mod 3). 2^2 is 4 = 1 (mod 3) of which we make a note (2^2 = 1 (mod 3)). Next, we find 0^1 = 0, 0^2 = 0 so we have a cycle of size 1 meaning no further multiplication is needed to tell 0^3 = 0 (mod 3). Note taken. Similar process for 1^4 (we tell on the second iteration that we have a cycle of size 1, so 1^4 is 1, which we note). Finally, we get 2^1 = 2 (mod 3), 2^2 = 1(mod 3), 2^3 = 2(mod 3), a cycle of length 2, so we can skip ahead an even number which exhausts 2^5 and without checking again we know that 2^5 = 2 (mod 3).
Our sum is now 1 + 1 + 1 + 0 + 1 + 2 (mod 3) = 2 + 1 + 0 + 1 + 2 (mod 3) = 0 + 0 + 1 + 2 (mod 3) = 0 + 1 + 2 (mod 3) = 1 + 2 (mod 3) = 0 (mod 3).
These rules will be helpful to you since your cases see N much larger than M. If this were reversed - if N were much smaller than M - you'd get no benefit from my method (and taking the modulus w.r.t. M would affect the outcome less).
Pseudocode:
Compute(N, M)
1. sum = 0
2. for i = 0 to N do
3. term = SelfPower(i, M)
4. sum = (sum + term) % M
5. return sum
SelfPower(k, M)
1. selfPower = 1
2. iterations = new HashTable
3. for i = 1 to k do
4. selfPower = (selfPower * (k % M)) % M
5. if iterations[selfPower] is defined
6. i = k - (k - i) % (i - iterations[selfPower])
7. clear out iterations
8. else iterations[selfPower] = i
9. return selfPower
Example execution:
resul = Compute(5, 3)
sum = 0
i = 0
term = SelfPower(0, 3)
selfPower = 1
iterations = []
// does not enter loop
return 1
sum = (0 + 1) % 3 = 1
i = 1
term = SelfPower(1, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 1 % 3) % 3 = 1
iterations[1] is not defined
iterations[1] = 1
return 1
sum = (1 + 1) % 3 = 2
i = 2
term = SelfPower(2, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 2 % 3) % 3 = 2
iterations[2] is not defined
iterations[2] = 1
i = 2
selfPower = (2 * 2 % 3) % 3 = 1
iterations[1] is not defined
iterations[1] = 2
return 1
sum = (2 + 1) % 3 = 0
i = 3
term = SelfPower(3, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 3 % 0) % 3 = 0
iterations[0] is not defined
iterations[0] = 1
i = 2
selfPower = (0 * 3 % 0) % 3 = 0
iterations[0] is defined as 1
i = 3 - (3 - 2) % (2 - 1) = 3
iterations is blank
return 0
sum = (0 + 0) % 3 = 0
i = 4
term = SelfPower(4, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 4 % 3) % 3 = 1
iterations[1] is not defined
iterations[1] = 1
i = 2
selfPower = (1 * 4 % 3) % 3 = 1
iterations[1] is defined as 1
i = 4 - (4 - 2) % (2 - 1) = 4
iterations is blank
return 1
sum = (0 + 1) % 3 = 1
i = 5
term = SelfPower(5, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 5 % 3) % 3 = 2
iterations[2] is not defined
iterations[2] = 1
i = 2
selfPower = (2 * 5 % 3) % 3 = 1
iterations[1] is not defined
iterations[1] = 2
i = 3
selfPower = (1 * 5 % 3) % 3 = 2
iterations[2] is defined as 1
i = 5 - (5 - 3) % (3 - 1) = 5
iterations is blank
return 2
sum = (1 + 2) % 3 = 0
return 0

Answer 2

why not just use simple recursion to find the recursive sum of the powers
def find_powersum(s):
if s == 1 or s== 0:
return 1
else:
return s*s + find_powersum(s-1)
def find_mod (s, m):
print(find_powersum(s) % m)
find_mod(4, 4)
2