ACM-ICPC 2018 徐州赛区网络预赛 D. EasyMath
ACM-ICPC 2018 徐州赛区网络预赛 D. EasyMath
做法:
\[f(m,n) = \sum _{i=1}^{m} \mu(in) = \sum_{i=1}^{m}[gcd(i,n)=1]\mu(i)\mu(n) = \mu(n)\sum_{d|n}\mu(d)f(\frac{m}{d},d)
\]
边界: n=1,杜教筛求\(\sum_{i=1}^{m}\mu(i)\),m = 1, 返回\(\mu(n)\),预处理尽可能把空间卡满。
2个小时的时候就推出来了这个式子,不会算复杂度,本校没人过。。。于是成功放弃了。。。
#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define mp make_pair
#define PII pair<int,int>
#define sc second
typedef long long ll;
const int N = 1e7 + 13000000 + 1;
const int LM = 13000000;
using namespace std;
ll n,m;
bool notp[N];
int p[N], smiu[N];
short miu[N];
void init() {
notp[1] = 1;
miu[1] = 1;
for(int i=2;i<=1e7+LM;++i) {
if(!notp[i]) p[++p[0]] = i, miu[i] = -1;
for(int j=1;j<=p[0]&&p[j]*i<=1e7+LM;++j) {
notp[i*p[j]] = 1;
if(i%p[j] == 0) {
miu[i*p[j]] = 0;
break;
}
miu[i*p[j]] = miu[i]*miu[p[j]];
}
}
for(int i=1;i<=1e7+LM;++i) smiu[i] = smiu[i-1] + miu[i];
}
ll g(ll n) {
if(n<=1e7+LM) return smiu[n];
if(n == 1) return 1;
ll ans = 1;
for(ll i=2,r;i<=n;i=r+1) {
r = (n/(n/i));
ans -= (r-i+1LL)*g(n/i);
}
return ans;
}
void chai(ll x,vector<ll> &v,ll &mu) {
v.clear();
mu = 1;
for(int i=1;i<=p[0]&&1LL*p[i]*p[i]<=x;++i) {
if(x%p[i]==0) {
int cnt = 0;
v.pb(p[i]);
mu = -mu;
while(x%p[i]==0) x/=p[i], cnt++;
if(cnt>1) mu = 0;
}
}
if(x!=1) mu=-mu, v.pb(x);
}
ll f(ll m,ll n) {
if(m == 0 || n == 0) return 0;
ll mu_n, ans = 0;
if(m == 1 && n <= 1e7+LM) return miu[n];
vector<ll> v;
chai(n,v,mu_n);
if(m == 1) return mu_n;
if(n == 1) return g(m);
if(mu_n == 0) return 0;
int cnt = v.size();
for(int s=0;s<(1<<cnt);++s) {
ll d = 1, mu_d = 1;
for(int i=0;i<cnt;++i) if(s&(1<<i)){
d = d*v[i];
mu_d = -mu_d;
}
if(m >= d) ans += mu_d*f(m/d,d);
}
ans *= mu_n;
return ans;
}
int main() {
init();
scanf("%lld%lld",&m,&n);
printf("%lld\n",f(m,n));
return 0;
}