You must first know what kind of number is "pure number of cycles."By meterCan be found in such numbers if and only if the denominator and \ (k \) are relatively prime, because, first of all consider the division process, each time give the remainder of the current \ (* k \) , and then do the division with remainder denominator, once the loop should appear to make a certain addition to the remainder after appear before too. there Euler's theorem and \ (a ^ {\ varphi ( n)} \ equiv 1 (\ mod n) (\ gcd (a, n ) = 1) \) , so it can make the remainder when a decimal point in the calculation reappears after several
Then to such different values, so they are actually required this thing \ [\ sum_ {i = 1 } ^ {n} \ sum_ {j = 1} ^ {m} [\ gcd (i, j) = 1] [\ gcd (j, k) = 1 ] \]
Only the first \ (J \) and \ (K \) ahead \ [\ sum_ {j = 1 } ^ {m} [\ gcd (j, k) = 1] \ sum_ {i = 1} ^ {n } [\ gcd (i, j ) = 1] \]
Then \ ([\ gcd (j, k) = 1] \) conversion click \ [\ sum_ {j = 1 } ^ {m} \ sum_ {d | j, d | k} \ mu (d) \ sum_ {i = 1} ^ {n } [\ gcd (i, j) = 1] \]
把\(d\)提前\[\sum_{d=1}^{m}\mu(d)\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor} \sum_{i=1}^{n}[\gcd(i,jd)=1]\]\[\sum_{d=1}^{m}\mu(d)\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor} \sum_{i=1}^{n}[\gcd(i,j)=1][\gcd(i,d)=1]\]
Behind that is a little familiar? If we remember \ (s (i, j, k) = \ sum_ {i = 1} ^ {n} \ sum_ {j = 1} ^ {m} [\ gcd (i, j ) =. 1] [\ GCD (J, K) =. 1] \) , then we can get \ [s (i, j, k) = \ sum_ {d | k} \ mu (d) s (\ lfloor \ frac {m} {d} \ rfloor, n, d) \]
This recursive process like, the boundary is \ (m = 0 \) or \ (n = 0 \) when \ (0 \) , \ (K =. 1 \) when \ (\ sum_ {i = 1 } ^ {n-} \ sum_ {J =. 1} ^ {m} [\ GCD (I, J) =. 1] \) , number Theory block + DU teach sieve can. complexity probably \ (O (\ log n \ log m \} + n-n-sqrt {^ {\ FRAC. 3} {2} {}) \) , this is relatively slow, the increase of the memory have more than 1800 ms
// luogu-judger-enable-o2
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<vector>
#include<cmath>
#include<ctime>
#include<queue>
#include<map>
#include<set>
#define LL long long
#define db double
using namespace std;
const int N=5e6+10,M=2000+10;
int rd()
{
int x=0,w=1;char ch=0;
while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
return x*w;
}
int n,m,k,prm[N],tt;
bool pp[N];
LL mu[N];
map<int,LL> f;
int lm;
LL siv(int nn)
{
if(nn<=N-10) return mu[nn];
if(f.count(nn)) return f[nn];
LL &an=f[nn];
an=1;
for(int i=2,j;i<=nn;i=j+1)
{
j=nn/(nn/i);
an-=1ll*siv(nn/i)*(j-i+1);
}
return an;
}
LL ff(int nn,int mm)
{
lm=min(nn,mm);
LL an=0;
for(int i=1,j;i<=lm;i=j+1)
{
j=min(nn/(nn/i),mm/(mm/i));
an+=1ll*(siv(j)-siv(i-1))*(nn/i)*(mm/i);
}
return an;
}
vector<int> dd[M];
struct node
{
int n,m,k;
bool operator < (const node &bb) const {return n!=bb.n?n<bb.n:(m!=bb.m?m<bb.m:k<bb.k);}
};
map<node,LL> g;
LL sov(int n,int m,int k)
{
if(n<=0||m<=0) return 0;
node nw=(node){n,m,k};
if(g.count(nw)) return g[nw];
if(k==1) return g[nw]=ff(n,m);
LL an=0;
vector<int>::iterator it;
for(it=dd[k].begin();it!=dd[k].end();++it)
{
int i=*it;
an+=1ll*sov(m/i,n,i)*(mu[i]-mu[i-1]);
}
return g[nw]=an;
}
int main()
{
mu[1]=1;
for(int i=2;i<=N-10;++i)
{
if(!pp[i]) prm[++tt]=i,mu[i]=-1;
for(int j=1;j<=tt&&i*prm[j]<=N-10;++j)
{
pp[i*prm[j]]=1;
if(i%prm[j]==0) break;
mu[i*prm[j]]=-mu[i];
}
}
for(int i=2;i<=N-10;++i) mu[i]+=mu[i-1];
n=rd(),m=rd(),k=rd();
for(int i=1;i<=k;++i)
if(k%i==0) dd[k].push_back(i);
int nn=dd[k].size();
for(int i=0;i<nn-1;++i)
{
int x=dd[k][i];
vector<int>::iterator it;
for(it=dd[k].begin();it!=dd[k].end();++it)
{
int y=*it;
if(x<y) break;
if(x%y==0) dd[x].push_back(y);
}
}
printf("%lld\n",sov(n,m,k));
return 0;
}