题目链接:luogu1587
首先是题目中的“纯循环小数”让人感觉十分清奇
对于一个分数(frac{x}{y}),要满足条件首先得使得(gcd(x,y)=1),其次因为纯循环必然存在一个循环节的长度(l)使得将小数点右移(l)位之后两个数的小数部分相同,写出来就是这个样子
[frac{x}{y}-lfloorfrac{x}{y}
floor=frac{xk^l}{y}-lfloorfrac{xk^l}{y}
floor
]
两边去分母可得
[x-ylfloorfrac{x}{y}
floor=xk^l-ylfloorfrac{xk^l}{y}
floor
]
移项
[xk^l-x=y(lfloorfrac{xk^l}{y}
floor-lfloorfrac{x}{y}
floor)
]
所以最后就是(xk^lequiv x(mod y)),由于(gcd(x,y)=1),所以两边同时除以(x)得到(k^lequiv 1(mod y)),很显然有(gcd(k,y)=1)
所以最后题目求的就是
[sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=1][gcd(j,k)=1]
]
如果两个都同时进行反演的话显然很难处理,于是考虑先做一个
[egin{aligned}
&sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=1][gcd(j,k)=1]\=&sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=1]sum_{d|gcd(j,k)}mu(d)\
=&sum_{d|k}mu(d)sum_{i=1}^nsum_{j=1}^{lfloorfrac{m}{d}
floor}[gcd(i,jd)=1]\
=&sum_{d|k}mu(d)sum_{i=1}^nsum_{j=1}^{lfloorfrac{m}{d}
floor}[gcd(i,j)=1][gcd(i,d)=1]
end{aligned}
]
这有什么意义呢?注意到前后两个式子形式大体一致,于是可以记(f(n,m,k)=sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=1][gcd(j,k)=1]),则(f(n,m,k)=sum_{d|k}mu(d)f(lfloorfrac{m}{d}
floor,n,d))
边界条件的话首先(f(0,m,k)=f(n,0,k)=0)
接下来(f(n,m,1)=sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=1]),这个随便反演一下就有(sum_{d=1}^nmu(i)lfloorfrac{n}{d}
floorlfloorfrac{m}{d}
floor),直接数论分块,但是要注意(nleq 10^9),于是考虑杜教筛求(mu)的和
一个小技巧就是我们只会去计算(mu(d)
eq 0)时的值,其它的直接剪枝即可
#include<iostream>
#include<string>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define maxd 1000000007
#define eps 1e-8
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
const int N=10000000;
struct node{
int n,m,k;
};
bool operator <(node p,node q)
{
if (p.n!=q.n) return p.n<q.n;
else if (p.m!=q.m) return p.m<q.m;
else return p.k<q.k;
}
map<node,ll> ans;
map<int,ll> sumu;
int n,m,k,tot=0,mu[N+10],pri[1001000],len;
bool nopri[N+10];
ll sum[N+10];
vector<int> fac;
int read()
{
int x=0,f=1;char ch=getchar();
while ((ch<'0') && (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
void sieve()
{
mu[1]=1;
rep(i,2,N)
{
if (!nopri[i]) {pri[++tot]=i;mu[i]=-1;}
int j;
for (j=1;j<=tot && i*pri[j]<=N;j++)
{
nopri[i*pri[j]]=1;
if (i%pri[j]==0) break;
else mu[i*pri[j]]-=mu[i];
}
}
rep(i,1,N) sum[i]=sum[i-1]+mu[i];
}
ll query(int x)
{
if (x<=N) return sum[x];
if (sumu[x]) return sumu[x];
ll ans=1;int l,r;
for (l=2;l<=x;l=r+1)
{
r=x/(x/l);
ans-=1ll*(r-l+1)*query(x/l);
}
sumu[x]=ans;
return ans;
}
ll calc(int n,int m)
{
if (n>m) swap(n,m);
//cout << n << " " << m << " ";
int l,r;ll ans=0;
for (l=1;l<=n;l=r+1)
{
r=min(n/(n/l),m/(m/l));
ans+=1ll*(n/l)*(m/l)*(query(r)-query(l-1));
}
//cout << ans << endl;
return ans;
}
ll solve(int n,int m,int k)
{
if ((!n) || (!m)) return 0;
node now=(node){n,m,k};
if (ans[now]) return ans[now];
if (k==1) ans[now]=calc(n,m);
else
{
int i;
for (i=0;i<len && fac[i]<=k;i++)
{
if ((k%fac[i]==0) && (mu[fac[i]]))
ans[now]+=solve(m/fac[i],n,fac[i])*mu[fac[i]];
}
}
return ans[now];
}
int main()
{
n=read();m=read();k=read();
sieve();
rep(i,1,k)
if (k%i==0) fac.pb(i);len=fac.size();
printf("%lld",solve(n,m,k));
return 0;
}