Returns the sum of all factors of a number.

For example, 6 has a factor of 1, 2, 3, and 6, and their sum is 12.
Divide the sum of all factors of N ^ m by the remainder of a number S. # Include < Cassert >
# Include < Vector >
Using   Namespace STD;

Static Vector < Int > Primes;
Bool Isprime ( Int N)
{
For (Size_t I =   0 ; Primes [I] * Primes [I] <= N; I ++ )
{
If(N%Primes [I]= 0)
Return False;
}
Return   True ;
}

Int Getprime (size_t index)
{
Assert (Index <   10000 );

If (Index > = Primes. Size ())
{
If (Primes. Size () =   0 )
{
Primes. Reserve ( 10000 );
Primes. push_back ( 2 );
Primes. push_back ( 3 );
Primes. push_back ( 5 );
Primes. push_back ( 7 );
}
For ( Int T = Primes. Back () +   2 ; Index > = Primes. Size (); t + =   2 )
{
If(Isprime (t ))
Primes. push_back (t );
}
}

Return Primes [Index];
}

_ Int64 powermod (_ int64 N, _ int64 M, _ int64 S)
{
_ Int64 T;
For (T =   1 ; M >>=   1 )
{
If (M &   1 )
T = T * N % S;
N = N * N % S;
}
Return T;
}

Int Sumfactormod ( Int N, Int M, Int S)
{
Int T;
Int Sum =   1 ;
For (T =   0 ; N >   1 ; T ++ )
{
Int P = Getprime (t );
Int Q =   0 ;
While (N % P =   0 )
N /= P, q ++ ;
If (Q)
{
/**/ /*
If n = PI {PI ^ Qi} pi is a prime number, sum can be expressed as pi {f (PI, Qi )}
Where F (p, q) = P ^ 0 + P ^ 1 + P ^ 2 + P ^ q = (P ^ (q + 1)-1)/P-1)
Therefore, the required value sum % s = PI {f (PI, Qi) % s} % s
F (p, q) % s = (P ^ (q + 1)-1)/(p-1) % s
Obtained by x/y % z = x % (y * Z)/y
F (p, q) % s = (P ^ (q + 1)-1) % (S * (PM)/(p-1)
= (P ^ (q + 1) % (S * (p-1) + (S * (PM)-1) % S * (p-1)/(p-1)
Among them, P ^ (q + 1) % (S * (PM) can be quickly obtained using powermod
*/
Q * = M;
_ Int64 sp_1 = _ Int64 (s) * (P -   1 );
_ Int64 t = Powermod (p, q + 1 , Sp_1 );
T + = Sp_1 -   1 ;
Sum * = T % Sp_1 / (P -   1 );
Sum % = S;
}
}
Return Sum % S;
}

Int Sumfactormod ( Int N, Int S)
{
ReturnSumfactormod (n,1, S );
}

Supplement: add the comment. Someone adds that if n = PI {PI ^ Qi} pi is a prime number, sum can be expressed as pi {f (PI, Qi )} "I don't understand. The following is a rough deduction:
N = PI {PI ^ Qi} pi is a prime number
Then the factor of N can be listed as follows: A1, A2,... a (PI {QI + 1:
A1 = 1 aq0 + 1 = p1... A (PI {QI + 1}-q0 + 1) = p1 ^ Q1 * P2 ^ Q2 *... * P0 ^ 0
A2 = P0 aq0 + 2 = p1 * P0... A (PI {QI + 1}-q0 + 2) = p1 ^ Q1 * P2 ^ Q2 *... * P0 ^ 1
A3 = P0 ^ 2 aq0 + 3 = p1 * P0 ^ 2... A (PI {QI + 1}-q0 + 3) = p1 ^ Q1 * P2 ^ Q2 *... * P0 ^ 2
.........
Aq0 = P0 ^ q0 aq0 + q0 = p1 * P0 ^ q0... A (PI {QI + 1}) = p1 ^ Q1 * P2 ^ Q2 *... * P0 ^ q0
We can see from the multiplication principle:
Among these factors, there are 0 PI factor factors: PI {QI + 1}/(QI + 1)
One PI factor and one pi {QI + 1}/(QI + 1)
...
There are Qi PI factors and PI {QI + 1}/(QI + 1)
This set of qi + 1 PI {QI + 1}/(QI + 1) numbers is exactly the same if the PI factor is removed separately.
This set is the factor set of the number N/PI ^ Qi after N removes the factor pi.
So g (n) = (PI ^ 0 + PI ^ 1 +... + PI ^ Qi) * g (N/PI ^ Qi)
That is, g (n) = PI {PI ^ 0 + PI ^ 1 +... + PI ^ Qi}