并不对劲的bzoj4652:loj2085:uoj221:p1587:[NOI2016]循环之美

题目大意

对于已知的十进制数\(n\)和\(m\)，在\(k\)进制下，有多少个数值上互不相等的纯循环小数，可以用\(x/y\)表示,其中 \(1\leq x\leq n,1\leq y\leq m\) (\(n,m\leq10^9,k\leq2000\))

题解

这个人（点这里）讲得很清楚\(\color{white}{\text{shing太强了}}\)

代码

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define maxn 2500010
#define lim 2500000
#define maxl 2010
#define LL long long
using namespace std;
int read()
{
	int x=0,f=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-')ch=getchar();
	if(ch=='-')f=-1,ch=getchar();
	while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	return x*f;
}
void write(LL x)
{
	if(x==0){putchar('0'),putchar('\n');return;}
	int f=0;char ch[20];
	if(x<0)putchar('-'),x=-x;
	while(x)ch[++f]=x%10+'0',x/=10;
	while(f)putchar(ch[f--]);
	putchar('\n');
	return;
}
int n,m,k,p[maxn],cnt,numk[100],num,f[maxn],no[maxn];
LL mu[maxn],ans;
map<int,LL>M;
map<pii,LL>G; 
int gcd(int x,int y){if(x>y)swap(x,y);if(!x)return y;return gcd(y%x,x);}
LL getm(int x)
{
	if(x<=lim)return mu[x];
	if(M[x])return M[x];
	LL res=1;
	for(int l=2,r=0;l<=x;l=r+1)r=x/(x/l),res-=(LL)(r-l+1)*(LL)getm(x/l);
	M[x]=res;
	return res;
}
LL g(int x,int y)
{
	if(!x)return getm(y);
	if(y<=1)return y;
	if(G[mp(x,y)])return G[mp(x,y)];
	return G[mp(x,y)]=g(x-1,y)+g(x,y/numk[x]);
}
int main()
{
	n=read(),m=read(),k=read();
	no[1]=mu[1]=1;
	rep(i,1,lim)
	{
		if(!no[i])mu[i]=-1,p[++cnt]=i;
		for(int j=1;j<=cnt&&p[j]*i<=lim;j++)
		{
			no[p[j]*i]=1;
			if(i%p[j]==0){mu[i*p[j]]=0;break;}
			else mu[i*p[j]]=-mu[i];
		}
	} 
	rep(i,2,lim)mu[i]+=mu[i-1];
	rep(i,1,k)f[i]=f[i-1]+(gcd(i,k)==1?1:0);
	for(int i=1;p[i]<=k;i++)if(k%p[i]==0)numk[++num]=p[i];
	for(int l=1,r=0;l<=min(n,m);l=r+1)r=min(n/(n/l),m/(m/l)),ans+=(LL)(g(num,r)-g(num,l-1))*(LL)(n/l)*(LL)(f[(m/l)%k]+(LL)((m/l)/k)*(LL)f[k]);
	write(ans);
	return 0;
}