Whitespace is a language only for polyglot. Such a use of this language is very strange. But I don’t know the appropriate language. piet, grass or forth?

## solution

Simplifying the code, finally it becomes like below. The answer is for $a = 100, b = 1000000000000$. http://ws2js.luilak.net/interpreter.html was useful.

#!/usr/bin/env python3
def sum_factors(x):
z = 0
for y in range(1, x+1):
if x % y == 0:
z += y
return z

a = int(input())
b = int(input())
c = -1
for i in range(a, b + 1):
print(i, sum_factors(i))
c = max(c, sum_factors(i))

print('TWCTF{%d}' % c)


Optimize it. The sum $5618427494400$ for $n = 995886571680$ is the answer. TWCTF{5618427494400}.

#include <iostream>
#include <vector>
#include <map>
#include <tuple>
#include <cmath>
#include <cassert>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
typedef long long ll;
using namespace std;
template <class T> void setmax(T & a, T const & b) { if (a < b) a = b; }
vector<int> sieve_of_eratosthenes(int n) { // enumerate primes in [2,n] with O(n log log n)
vector<bool> is_prime(n+1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i*i <= n; ++i)
if (is_prime[i])
for (int k = i+i; k <= n; k += i)
is_prime[k] = false;
vector<int> primes;
for (int i = 2; i <= n; ++i)
if (is_prime[i])
primes.push_back(i);
return primes;
}
map<ll,int> factors(ll n, vector<int> const & primes) {
map<ll,int> result;
for (int p : primes) {
if (n < p *(ll) p) break;
while (n % p == 0) {
result[p] += 1;
n /= p;
}
}
if (n != 1) result[n] += 1;
return result;
}

ll sum_factors(ll n, vector<int> const & primes) {
ll ans = 1;
for (auto it : factors(n, primes)) {
ll p; int cnt; tie(p, cnt) = it;
ll acc = 0; ll p_i = 1;
repeat (i,cnt+1) {
acc += p_i;
if (i != cnt /* to avoid overflow */) p_i *= p;
}
ans *= acc;
}
return ans;
}
pair<ll,ll> dfs(int i, ll n, ll acc, ll a, ll b, vector<int> const & primes) {
pair<ll,ll> result = { acc, n };
if (i >= 30) return result;
int k = 0;
ll pk = 1;
ll pkacc = 1;
for (; n * pk * primes[i] <= b; ++ k) {
pk *= primes[i];
pkacc += pk;
}
setmax(result, make_pair(acc * pkacc, n * pk));
for (; k >= 0; -- k) {
setmax(result, dfs(i+1, n * pk, acc * pkacc, a, b, primes));
pkacc -= pk;
pk /= primes[i];
}
return result;
}
int main(int argc, char **argv) {
assert (argc == 1);
ll a, b; cin >> a >> b;
vector<int> primes = sieve_of_eratosthenes(sqrt(b) + 3);
auto it = dfs(0, 1, 1, a, b, primes);
cout << it.first << ' ' << it.second << endl;
return 0;
}