E. Present for Vitalik the Philatelist 反演+容斥

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题意:给n个数\(a_i\),求选一个数x和一个集合S不重合,gcd(S)!=1,gcd(S,x)==1的方案数.
题解:\(ans=\sum_{i=2}^nf_ig_i\),\(f_i\)是数组中和i的gcd不为1的个数,\(g_i\)是选取集合gcd为i的方案数.
\(f_n=\sum_{i=1}^N[gcd(n,i)!=1]a_i\)
\(f_n=\sum_{i=1}^N\sum_{d|gcd(i,n)}\mu(d)a_i\)
\(f_n=\sum_{d|n}\mu(d)\sum_{i=1}^{\frac{N}{d}}a_{i*d}\)
\(f\)可以\(nlogn\)预处理
\(g_i=2^{b_i}-1-\sum_{i|x}g(x)\),\(b_i\)是数组中是i倍数的数的个数,可以\(nlogn\)处理

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
//#include <bits/extc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define mt make_tuple
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
#define bpc __builtin_popcount
#define base 1000000000000000000ll
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
#define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;
//using namespace __gnu_pbds;

const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=10000000+10,maxn=500000+10,inf=0x3f3f3f3f;

struct FastIO {
    static const int S = 1e7;
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) {}
    inline int xchar() {
        static char buf[S];
        static int len = 0, pos = 0;
        if (pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if (pos == len) exit(0);
        return buf[pos++];
    }
    inline int xuint() {
        int c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x;
    }
    inline int xint()
    {
        int s = 1, c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        if (c == '-') s = -1, c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while (c <= 32) c = xchar();
        for (; c > 32; c = xchar()) * s++ = c;
        *s = 0;
    }
    inline void wchar(int x)
    {
        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos++] = x;
    }
    inline void wint(ll x)
    {
        if (x < 0) wchar('-'), x = -x;
        char s[24];
        int n = 0;
        while (x || !n) s[n++] = '0' + x % 10, x /= 10;
        while (n--) wchar(s[n]);
        wchar('\n');
    }
    inline void wstring(const char *s)
    {
        while (*s) wchar(*s++);
    }
    ~FastIO()
    {
        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;
int mi[maxn],b[N],c[N],a[N],mu[N];
int main()
{
    int n=io.xint(),ma=0;
    mi[0]=1;
    for(int i=1;i<=n;i++)
    {
        int x=io.xint();
        b[x]++;ma=max(ma,x);
        mi[i]=(1ll*mi[i-1]<<1)%mod;
    }
    mu[1]=1;
    for(int i=1;i<=ma;i++)
    {
        for(int j=i<<1;j<=ma;j+=i)
            b[i]+=b[j],mu[j]-=mu[i];
        for(int j=i;j<=ma;j+=i)
            c[j]+=mu[i]*b[i];
    }
    for(int i=ma;i;i--)
    {
        a[i]+=mi[b[i]]-1+mod;
        if(a[i]>=mod)a[i]-=mod;
        for(int j=i<<1;j<=ma;j+=i)
        {
            a[i]-=a[j];
            if(a[i]<0)a[i]+=mod;
        }
    }
    ll ans=0;
    for(int i=2;i<=ma;i++)if(c[i]&&a[i])add(ans,1ll*c[i]*a[i]%mod);
    io.wint(ans);
    return 0;
}
/********************
1
10000000
********************/

E. Present for Vitalik the Philatelist 反演+容斥

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原文地址：https://www.cnblogs.com/acjiumeng/p/11129181.html