题目链接
题解
orz
此题太优美了
我们令(frac{x}{y})为最简分数，则(x perp y)即，(gcd(x,y) = 1)
先不管(k)进制，我们知道(10)进制下如果(frac{x}{y})是纯循环的，只要(2 perp y)且(5 perp y)
可以猜想在(k)进制下同样成立
证明：
若(frac{x}{y})为纯循环小数，设其循环节长度为(l)，那么一定满足
[{ frac{xk^{l}}{y} } = { frac{x}{y}}
]
其中({x})指小数部分
可以写成
[xk^{l}- lfloor frac{xk^{l}}{y}
floor * y= x - lfloor frac{x}{y}
floor * y
]
可以看出就是同余的形式
[xk^{l} equiv x pmod y
]
由于(x perp y)
[k^{l} equiv 1 pmod y
]
要使存在(l)，使得式子成立，那么一定(k perp y)
所以我们有
[egin{aligned}
ans &= sumlimits_{i = 1}^{n} sumlimits_{j = 1}^{m} [i perp j] [j perp k] \
&= sumlimits_{j = 1}^{m} [j perp k] sumlimits_{i = 1}^{n} [(i,j) == 1] \
&= sumlimits_{j = 1}^{m} [j perp k] sumlimits_{i = 1}^{n} epsilon((i,j)) \
&= sumlimits_{j = 1}^{m} [j perp k] sumlimits_{i = 1}^{n} sumlimits_{d|(i,j)} mu(d) \
&= sumlimits_{d = 1}^{n} mu(d) sumlimits_{d|i}^{n} sumlimits_{d|j}^{m} [j perp k] \
&= sumlimits_{d = 1}^{n} [d perp k] mu(d) lfloor frac{n}{d}
floor sumlimits_{j}^{lfloor frac{m}{d}
floor} [j perp k] \
end{aligned}
]
我们令
[f(n) = sumlimits_{i}^{n} [i perp k]
]
根据(gcd(i + k,k) = gcd(i,k))，我们有
[f(n) = lfloor frac{n}{k}
floor f(k) + f(n mod k)
]
所以我们只需要(O(klogk))暴力计算(f(1....k))就可以(O(1))计算后面的式子了
现在考虑前面的
[sumlimits_{d = 1}^{n} [d perp k] mu(d)
]
为了使能整除分块，我们必须算出其前缀和
令
[g(n,k) = sumlimits_{i = 1}^{n} [i perp k] mu(d)
]
用类似上面同样的方法：
[egin{aligned}
g(n,k) &= sumlimits_{i = 1}^{n} [i perp k] mu(i) \
&= sumlimits_{i = 1}^{n} mu(i) sumlimits_{d|i,d|k} mu(d) \
&= sumlimits_{d | k} mu(d) sumlimits_{d|i} mu(i) \
&= sumlimits_{d | k} mu(d) sumlimits_{i = 1}^{lfloor frac{n}{d}
floor} mu(id) \
&= sumlimits_{d | k} mu(d) sumlimits_{i = 1}^{lfloor frac{n}{d}
floor} [i perp d]mu(i) * mu(d) \
&= sumlimits_{d | k} mu(d)^2 sumlimits_{i = 1}^{lfloor frac{n}{d}
floor} [i perp d]mu(i) \
&= sumlimits_{d | k} mu(d)^2 g(lfloor frac{n}{d}
floor,d) \
end{aligned}
]
就可以递归求解
当(n = 0)时，(g(0,k) = 0)
当(k = 1)时，(g(n,1) = sumlimits_{i = 1}^{n} mu(i))，上杜教筛即可
因为(lfloor frac{n}{d}
floor)只有(sqrt{n})种取值，(d)只能取(k)的因子，记其数量为(p)
那么求(g(n,k))总的复杂度为(O(psqrt{n} + n^{frac{2}{3}}))
于是乎我们就解决这道题了
总复杂度(O(psqrt{n} + n^{frac{2}{3}} + sqrt{n} + klogk))
#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
#define mp(a,b) make_pair<LL,LL>(a,b)
using namespace std;
const int maxn = 1000005,maxm = 100005,INF = 1000000000;
map<LL,LL> _mu;
map<LL,LL>::iterator it;
map<pair<LL,LL>,LL> G;
map<pair<LL,LL>,LL>::iterator IT;
int N;
int p[maxn],pi,isn[maxn];
LL Smu[maxn],f[maxn],mu[maxn];
LL n,m,K;
int gcd(int a,int b){return b ? gcd(b,a % b) : a;}
void init(){
N = 1000000;
mu[1] = 1;
for (int i = 2; i <= N; i++){
if (!isn[i]) p[++pi] = i,mu[i] = -1;
for (int j = 1; j <= pi && i * p[j] <= N; j++){
isn[i * p[j]] = true;
if (i % p[j] == 0){
mu[i * p[j]] = 0;
break;
}
mu[i * p[j]] = -mu[i];
}
}
for (int i = 1; i <= N; i++) Smu[i] = Smu[i - 1] + mu[i];
for (int i = 1; i <= K; i++) f[i] = f[i - 1] + (gcd(i,K) == 1);
}
LL S(LL n){
if (n <= N) return Smu[n];
if ((it = _mu.find(n)) != _mu.end()) return it->second;
LL ans = 1;
for (LL i = 2,nxt; i <= n; i = nxt + 1){
nxt = n / (n / i);
ans -= (nxt - i + 1) * S(n / i);
}
return _mu[n] = ans;
}
LL F(LL n){
return (n / K) * f[K] + f[n % K];
}
LL g(LL n,LL k){
if ((IT = G.find(mp(n,k))) != G.end())
return IT->second;
if (n == 0) return 0;
if (k == 1) return S(n);
LL ans = 0;
for (LL i = 1; i * i <= k; i++){
if (k % i == 0){
if (mu[i]) ans += g(n / i,i);
if (i * i != k && mu[k / i])
ans += g(n / (k / i),k / i);
}
}
return G[mp(n,k)] = ans;
}
int main(){
cin >> n >> m >> K;
init();
LL ans = 0,now,last = 0;
for (LL i = 1,nxt; i <= min(n,m); i = nxt + 1){
nxt = min(n / (n / i),m / (m / i));
now = g(nxt,K);
ans += 1ll * (now - last) * (n / i) * F(m / i);
last = now;
}
cout << ans << endl;
return 0;
}