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  3. Project Euler #48: Self powers
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Project Euler #48: Self powers

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66 Discussions

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  • samitdebnath435
     + 1 comment

    def Square_sum(n): MOD = 10**10 

    total_sum = 0


for i in range(1, n + 1):

    total_sum = (total_sum + pow(i, i, MOD)) % MOD

return total_sum

    if name == 'main':

    N = int(input())


print(Square_sum(N))

    • samitdebnath435
       + 0 comments

      def Square_sum(n):

      variable = []

for i in range(1 , n+1):

    square = i**i
    variable.append(square)



result = sum(variable)

last_10_digits = result % (10**10)

return last_10_digits

      if name == 'main':

      N = int(input())


print(Square_sum(N))

  • hamblinglen
     + 1 comment

    Python 3 100% 1-liner

    print(int(str(sum(pow(i, i, 10**10) for i in range(1, int(input())+1)))[-10:]))

    • hamblinglen
       + 0 comments

      The above is not the most computationally efficient.

      If you don't count import statements againsed the "1-liner" qualifier, the following is better:

      ` from functools import reduce

      print(reduce(lambda a, b: (a+b) % (10**10), [pow(i, i, 10**10) for i in range(1, int(input())+1)])) `

  • srikv
     + 0 comments

    python3 100%

    # Enter your code here. Read input from STDIN. Print output to STDOUT

N = int(input())

val, e = 0, 10**10

def exponent_mod(a, n, e):
    p = 1
    while n > 1:
        if n % 2 == 1:
            p *= a % e
            n = (n-1)
        a = (a**2) % e
        n //= 2
    return((p*a) % e)
    
for i in range(1, N+1):
    #print(exponent_mod(i, i , e), pow(i, i, e))
    val += exponent_mod(i, i, e)
    
print(val%e)

  • ketan_patel25
     + 0 comments

    JAva code

    import java.io.*;
import java.util.*;
import java.math.BigInteger;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long t = in.nextLong();
        System.out.println(count(t));
    }


    public static String count(long num) {
        BigInteger modulus = BigInteger.TEN.pow(10);
        BigInteger sum = BigInteger.ZERO;
        for (int i = 1; i <= num; i++)
            sum = sum.add(BigInteger.valueOf(i).modPow(BigInteger.valueOf(i), modulus));
        return sum.mod(modulus).toString();
    }
}

  • nc1969
     + 0 comments

    I was pretty much surprised that my solution has been accepted 100 % while it took 3.5 seconds on my (modest though) laptop. I thought the correct solutions should take 2 sec at most.

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