Approach and Code for o(log n) solution

问题

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)

回答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

回答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

来源：https://stackoverflow.com/questions/45175610/approach-and-code-for-olog-n-solution

标签

algorithm

time-complexity

number-theory