[Number Theory] [enumeration approx.] [Euler's function] bzoj2705 [sdoi2012] longge Problem

∵ Σ gcd (I, n) (1 <= I <= N)

= K1 * s (F1) + K2 * s (K2) +... + km * s (Km) {Ki is the approximate number of N, and S (Ki) is the condition that gcd (x, n) = KI (1 <= x <= N) number of x}

Export gcd (x, n) = KI (1 <= x <= N) <=> gcd (x/ki, N/ki) = 1 (1 <= x/ki <= N/ki)

The number of X in gcd (x/ki, N/ki) = 1 (1 <= x/ki <= N/ki) is)

Radians = Σ PHI (N/ki) * ki

∴ O (SQRT (N) enumerative approx.

 1 #include<cstdio> 2 #include<cmath> 3 using namespace std; 4 typedef long long ll; 5 ll n,ans; 6 int phi(ll x) 7 { 8     ll res=x; 9     for(ll i=2;i*i<=x;i++)10       if(x%i==0)11         {12           res=res/i*(i-1);13           while(x%i==0) x/=i;14         }15     if(x>1) res=res/x*(x-1);16     return res;17 }18 int main()19 {20     scanf("%lld",&n);21     for(ll i=1;i*i<=n;i++) if(n%i==0) ans+=(phi(n/i)*i+phi(i)*(n/i));22     printf("%lld\n",ans);23     return 0;24 }

 

[Number Theory] [enumeration approx.] [Euler's function] bzoj2705 [sdoi2012] longge Problem