# [ACCEPTED]-Sum of series: 1^1 + 2^2 + 3^3 + ... + n^n (mod m)-series

Score: 24

Two notes:

``````(a + b + c) % m
``````

is equivalent to

``````(a % m + b % m + c % m) % m
``````

and

``````(a * b * c) % m
``````

is equivalent 4 to

``````((a % m) * (b % m) * (c % m)) % m
``````

As a result, you can calculate each term 3 using a recursive function in O(log p):

``````int expmod(int n, int p, int m) {
if (p == 0) return 1;
int nm = n % m;
long long r = expmod(nm, p / 2, m);
r = (r * r) % m;
if (p % 2 == 0) return r;
return (r * nm) % m;
}
``````

And 2 sum elements using a `for` loop:

``````long long r = 0;
for (int i = 1; i <= n; ++i)
r = (r + expmod(i, i, m)) % m;
``````

This algorithm 1 is O(n log n).

Score: 5

I think you can use Euler's theorem to avoid 3 some exponentation, as phi(200000)=80000. Chinese 2 remainder theorem might also help as it 1 reduces the modulo.

Score: 3

You may have a look at my answer to this post. The 5 implementation there is slightly buggy, but 4 the idea is there. The key strategy is to 3 find x such that n^(x-1)<m and n^x>m 2 and repeatedly reduce n^n%m to (n^x%m)^(n/x)*n^(n%x)%m. I 1 am sure this strategy works.

Score: 1

I encountered similar question recently: my 3 'n' is 1435, 'm' is 10^10. Here is my solution 2 (C#):

``````ulong n = 1435, s = 0, mod = 0;
mod = ulong.Parse(Math.Pow(10, 10).ToString());
for (ulong i = 1; i <= n;
{
ulong summand = i;
for (ulong j = 2; j <= i; j++)
{
summand *= i;
summand = summand % mod;
}
s += summand;
s = s % mod;
}
``````

At the end 's' is equal to required 1 number.

Score: 0

Are you getting killed here:

``````for(j=1;j<=i;j++)
t=((long long)t*i)%m;
``````

Exponentials 2 mod m could be implemented using the sum 1 of squares method.

``````n = 10000;
m = 20000;
sqr = n;
bit = n;
sum = 0;

while(bit > 0)
{
if(bit % 2 == 1)
{
sum += sqr;
}
sqr = (sqr * sqr) % m;
bit >>= 2;
}
``````
Score: 0