Project Euler 516 5-smooth totients (数论)

题目链接：

https://projecteuler.net/problem=516

题目：

(5)-smooth numbers are numbers whose largest prime factor doesn't exceed (5).

(5)-smooth numbers are also called Hamming numbers.

Let S(L) be the sum of the numbers (n) not exceeding (L) such that Euler's totient function (φ(n)) is a Hamming number.

S((100))=(3728).

Find S((10^{12})). Give your answer modulo (2^{32}).

题解：

因为：

• • 如果 (n) 是质数，则(φ(n)=n-1)

• • 如果 (n) 是质数，则(φ(n^k) = n^k *(n-1))

• • 如果 (x) 和 (y) 互质，则(φ(xy) = φ(x) * φ(y))

而且，(5)-smooth number 可以表示成 (2^a 3^b 5^c)

那么我们可以容易得到：

题目要求的就是：(n = 2^a 3^b 5^c cdot p_1 cdot p_2 cdot dotso cdot p_k)。

其中，(p_i = 2^{a_i} 3^{b_i} 5^{c_i} + 1)。

直接预处理所有 (10^{12}) 内的 (p_i)，然后暴搜所有可能的乘积。

把这些可能的结果相加起来就是答案。

代码：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e8;
const int mod = 1e9;

const ll limit = 1e12;

int isprime(ll x)
{
  if(x<=1)return 0;
  for(ll i = 2;i * i <= x; i++) {
    if(x % i == 0) {
      return 0;
    }
  }
  return 1;
}

ll ans = 0;
std::vector<ll> v;
// n = 2^a * 3^b * 5^c * p_1 *p_2*...*p_k 
// where p_i = 2^a_i * 3^b_i * 5^ c_i + 1
//generate all the possible products and sum them
void dfs(ll id,ll now,ll upper)
{
// (now) value is a part of products
 if(v[id] > upper || id == v.size()) {
  //  std::cout << "now="<< now << '
';
   //ll sum = 0;
   // if(lim==1e12) {
   //   std::cout << "id=" << id << '
'; // 546
   // }
    for(ll x = now ; x <= limit ; x *= 2LL) {
      for(ll y = 1 ; x*y <= limit ; y *= 3LL) {
        for(ll z = 1 ; x*y*z <= limit; z *= 5LL) {
           ans += (int)(x*y*z); //a little bug , need to change ll into interger
      //     sum += (int)(x*y*z);
      //   std::cout << "cal=" << (x*y*z) << '
';
        }
      }
    }
  //  std::cout <<"sum=" << sum << '
';
    return;
  }
  dfs(id + 1, now, upper);
  dfs(id + 1, now * v[id], upper / v[id]);
}
int main(int argc, char const *argv[]) {

  for(ll x = 1; x <= limit;  x *= 2LL) {
    for(ll y = 1; x*y <= limit; y *= 3LL) {
      for(ll z = 1 ; x*y*z <= limit; z *= 5LL) {
        // if n is a prime, call it "good" prime
        // phi(n) = n - 1 = 2 ^ k_1 * 3^k_2 * 5^k_3
        // ==> n = 2 ^ k_1 * 3^k_2 * 5^k_3 + 1
        if(isprime(x*y*z + 1)) {
          // 2 ^ k_1 * 3^k_2 * 5^k_3 + 1
          v.push_back(1LL*(x*y*z + 1));
        }
      }
    }
  }
  sort(v.begin(),v.end());

  // std::cout << "size=" << v.size() << '
';
  // std::cout << "finish!" << '
';
  // std::cout  << '
';

 // 3LL means that  Skip 2, 3, 5
  dfs(3LL,1LL,limit);
  ans = ans % (1LL<<32);
  std::cout << ans << '
';
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";
  return 0;
}