Problem 288: An enormous factorial

(see projecteuler.net/problem=288)

For any prime p the number N(p,q) is defined by N(p,q) = sum_{n=0 ... q}{T_n * p^n}
with T_n generated by the following random number generator:

S_0 = 290797
S_{n+1} = S_n^2 mod 50515093
T_n = S_n mod p

Let Nfac(p,q) be the factorial of N(p,q).
Let NF(p,q) be the number of factors p in Nfac(p,q).

You are given that NF(3,10000) mod 3^20 = 624955285.

Find NF(61, 10^7) mod 61^10

My Algorithm

I started with the easy part:
The function countFactors() returns the exponent of a given prime in a factorial's prime factorization.
It relies on Legendre's formula which I already used for other problems, too ( see en.wikipedia.org/wiki/Legendre's_formula).

The pseudo-random number generator wasn't hard as well. I wrote that stuff when I saw the problem for the first time but then I was stuck and continued with other problems.
Upen revisiting this problem I made the crucial observation that the result has to be printed mod 61^10 → and 61 is the same prime used for NF(61, 10^7).
(the same hold true for the example NF(3, 10000) mod 3^20).

Therefore the result depends on three parameters:

The term p^n in the formula N(p,q) = sum_{n=0 ... q}{T_n * p^n} will be p=61 and n=0,1,... therefore 61^0 = 1, 61^1 = 61, 61^2 = 3721, ...
It can't exceed 61^10 because of the mod 61^10 part in the final formula. The variable maxPower will be 1, 61, 3721, ... until it reaches 61^10 and remains at the level.

There are only 61 possible values for T_n = S_n mod 61. That means that the number of different T_n * p^n is very small (in fact it's 71).
Caching those results of countFactors(t * p^n) let's the runtime of the program drop from 2.6 to 0.35 seconds.

Note

T_n is at most 60, so max(T_n * 61^10) = 60 * 61^10 approx 4.28 * 10^19 is the largest value I will encounter in my computations.
Unfortunately this value is beyond the range of 64 bit integers (it requires 66 bits) and I needed to switch to GCC's 128 bit extension.
Therefore this code doesn't compile with Visual C++ etc.

I didn't like this problem because I spent too much time "discovering" that I need 128 bit arithmetic.

Interactive test

This feature is not available for the current problem.

My code

… was written in C++11 and can be compiled with G++. You can download it, too.

#include <iostream>
#include <vector>
#include <unordered_map>
 
// I had overflows with 64 bit arithmetic, so I switched to GCC's 128 bit extension
typedef __int128 Number;
 
// compute the exponent of prime in the factorization of f!
unsigned long long countFactors(Number f, unsigned int prime)
{
// Legendre's formula, https://en.wikipedia.org/wiki/Legendre%27s_formula
unsigned long long result = 0;
Number power = prime;
while (power <= f)
{
result += f / power;
power *= prime;
}
 
return result;
}
 
// variables of the problem statement:
// p = prime = 61
// q = iterations = 10^7
// moduloExponent = 10 (because of 61^10)
unsigned long long solve(unsigned int prime, unsigned int iterations, unsigned int exponent)
{
// compute prime^moduloExponent => 61^10
unsigned long long modulo = 1;
for (unsigned int i = 1; i <= exponent; i++)
modulo *= prime;
 
// will be 61^i => 61^0, 61^1, 61^2, ..., 61^10 and keep it at 61^10
Number maxPower = 1;
 
// seed of pseudo-random number generator
unsigned long long s = 290797;
 
unsigned long long result = 0;
for (unsigned int i = 0; i <= iterations; i++)
{
// T(i, 61) * 61^i
auto t = s % prime;
// => but 61^i limited to 61^10
auto product = t * maxPower;
 
// I observed only 71 different products (in 1000000 iterations !), let's cache them
static std::unordered_map<Number, unsigned long long> cache;
auto lookup = cache.find(product);
if (lookup == cache.end())
{
// new, unknown parameters, must call countFactors()
auto current = countFactors(product, prime);
// cache the value
cache[product] = current;
// add it to the result, too
result += current;
}
else
{
// add looked up value
result += lookup->second;
}
 
// keep under 61^10
result %= modulo;
 
// 61^i => 61^(i+1), but limit to 61^10
if (maxPower < modulo)
maxPower *= prime;
 
// next random number
s *= s;
s %= 50515093;
}
 
return result;
}
 
int main()
{
unsigned int prime = 61;
unsigned int iterations = 10000000;
unsigned int exponent = 10;
std::cin >> prime >> iterations >> exponent;
std::cout << solve(prime, iterations, exponent) << std::endl;
return 0;
}

This solution contains 13 empty lines, 20 comments and 3 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.3 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

Changelog

October 19, 2017 submitted solution
October 19, 2017 added comments

Difficulty

35% Project Euler ranks this problem at 35% (out of 100%).

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