题目链接：

https://projecteuler.net/problem=429

题目：

A unitary divisor \(d\) of a number \(n\) is a divisor of \(n\) that has the property \(gcd(d, n/d) = 1\).

The unitary divisors of \(4! = 24\) are \(1, 3, 8\) and \(24\).

The sum of their squares is \(1^2 + 3^2 + 8^2 + 24^2 = 650\).

Let \(S(n)\) represent the sum of the squares of the unitary divisors of \(n\). Thus \(S(4!)=650\).

Find \(S(100 000 000!)\) modulo \(1 000 000 009\).

题解：

因为：\(n! = p_1^{a_1}p_2^{a_2}p_3^{a_3} \cdots p_k^{a_k}\)

所以：\(S(n!) = S(p_1^{a_1}p_2^{a_2}p_3^{a_3} \cdots p_k^{a_k}) = S(p_1^{a_1})*S(p_2^{a_2})*\cdots*S(p_k^{a_k})\)

\(=\displaystyle \prod_{i=1}^k (p_i^{2a_i}+1)\)

其实，unitary divisor 就是 \(1, p_1^{a_1}, p_2^{a_2}, p_3^{a_3}, \cdots p_k^{a_k}\)

比如: \((1 + a)(1 + b)(1 + c) = 1 + a + b + c + ab + ac + bc + abc\)

所以，他们的和就是 \(\displaystyle \prod_{i=1}^k (p_i^{a_i}+1)\)。

同理，它们的平方和就是 \(\displaystyle \prod_{i=1}^k (p_i^{2a_i}+1) = \displaystyle \prod_{i=1}^k ((p_i^{a_i})^{2}+1)\)

其中，\(\displaystyle a_i =\sum_{j=1}^{ n } \left\lfloor \frac{n}{p_i^j} \right\rfloor\)

代码：

#include <bits/stdc++.h>using namespace std;
typedef long long ll;
const int maxn = 1e8;
const int mod =1e9+9;ll qpower(ll a,ll b,ll mod)
{long long ans=1;while(b>0){if(b&1)ans=(ans*a)%mod;b>>=1;a=(a*a)%mod;}return ans;
}ll solve(ll i,ll n)
{ll res = 0;while (n) {n /= i;res += n;}
//  std::cout << "res=" << res << '\n';return res;
}ll cal(ll a)
{return (1LL + a * a) % mod;
}int main(int argc, char const *argv[]) {ll ans = 1;std::vector<ll> isprime(1e8+123,1);isprime[0] = 0;isprime[1] = 0;isprime[2] = 1;for(int i=2;i<maxn;i++) {if(isprime[i]) {for(int j= i+i;j<maxn;j+=i) {isprime[j] = 0;}}}
//  std::cout << "init finish" << '\n';for(ll i=2;i<maxn;i++) {//  if(i%10000000==0)std::cout << "ok" << '\n';if(isprime[i]) {ll power = solve(i,maxn);ans = 1LL * ans * cal( qpower(i,power,mod) ) % mod;}}std::cout << ans << '\n';cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";return 0;
}