ACM-ICPC 2018 徐州赛区网络预赛 D. EasyMath

Posted 2021-01-05 rrrr-wys

ACM-ICPC 2018 徐州赛区网络预赛 D. EasyMath

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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
",f(m,n));
    return 0;
}