UVa1363

  This question is a solution to a problem from the purple book, here to write about the deepening impact

  Gives the N, M satisfying M <= N is obtained 2-N! Present in all x such that x number of prime factors greater than M, then all prime factors greater than M and M is equivalent to! Relatively prime, Euclidean algorithm,

We x and M! Relatively prime converted to x% M! M and! Relatively prime, so we will only need to enumerate smaller than M! Number, then the number obtained * N! / M! (Mean increase in number a M! still satisfy the condition), to obtain less than M! M and the number! relatively prime, we use is the Euler function , and that we can find phifac (n with phifac (n) = phi (n !)) relationship phifac (n-1) with their prime factors tend to be the same, if n is not a prime number, phifac (n) = phifac ( n-1) * n otherwise phifac (n) = phifac (n -1) * ( n-1)

This resulted in the size phifac (n) of the note overflow, I am not particularly clear some of the details, here's the code:

// UVa 11440
#include <cstdio> 
#include <cstring>
#include <cmath> 
using namespace std; 

const int maxn = 10000000 + 5; 
const int MOD = 100000007; 

int vis[maxn], phifac[maxn];

void Eratosthenes(int n) {
    memset(vis, 0, sizeof(vis)); 
    int m = sqrt(n + 0.5); 
    for (int i = 2; i <= m; ++i) 
      for (int j = i*i; j <= n; j += i) 
        vis[j] = 1;
}

int main() { 
  int n, m; 
  Eratosthenes(10000001); 
  phifac[1] = phifac[2] = 1; 
  for (int i = 3; i <= 10000001; ++i) 
    phifac[i] = (long long)phifac[i-1] * (vis[i] ? i : i-1) % MOD;
    
  while (scanf("%d%d", &n, &m) == 2 && n) {    
    int ans = phifac[m]; 
    for (int i = m+1; i <= n; ++i) ans = (long long)ans * i % MOD; 
    printf("%d\n", (ans-1+MOD)%MOD); 
  }
  return 0; 
}