模板——数论

线性逆元

	inv[1]=1;
    for(int i=2;i<=n;++i){
        inv[i]=prd((P-P/i),inv[P%i]);
    }

线性筛

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void A(int& x,int y){
	x+=y;
	if(x>=P) x-=P;
}
int u[N],s[N],pri[N];
bool vis[N];
vector<int>d[N];
void init(){
	int n=100000,tot=0;
	u[1]=1;
	for(int i=2;i<=n;i++){
		if(!vis[i]){
			pri[++tot]=i;
			u[i]=P-1;
		}
		for(int j=1;j<=tot && pri[j]*i<=n;j++){
			int x=pri[j]*i;
			vis[x]=1;
			if(!(i%pri[j])){
				u[x]=0;
				break;
			}
			u[x]=P-u[i];
		}
	}
	for(int i=1;i<=n;i++){
		A(s[i],u[i]);
		d[i].push_back(i);
		//if(i<=10) cout<<i<<" "<<u[i]<<" "<<s[i]<<endl;
		for(int j=2;i*j<=n;j++){
			A(s[i*j],1ll*u[i]*j%P);
			d[i*j].push_back(i);
		}
	}
}

min-25筛求质数+数论分块

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#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1000010,P=1e9+7;

int prime[N], id1[N], id2[N], flag[N], ncnt, m;
ll g[N], sum[N], a[N], T;
ll n;

int ID(ll x){
    return x <= T ? id1[x] : id2[n / x];
}

ll calc(ll x){
    return (x%P * ( (x + 1)%P ) %P*((P+1)/2)%P +P- 1)%P;
}

ll f(ll x){
    return x;
}

void init(ll n){
    T = sqrt(n + 0.5);
    for(int i = 2; i <= T; i ++){
        if(!flag[i]) prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;
        for(int j = 1;j <= ncnt && i * prime[j] <= T; j ++){
            flag[i * prime[j]] = 1;
            if(i % prime[j] == 0) break; 
        }
    }
    for(ll l = 1; l <= n; l = n / (n / l) + 1){
        a[++m] = n / l;
        if(a[m] <= T) id1[a[m]] = m; else id2[n / a[m]] = m;
        g[m] = calc(a[m]);
    }
    for(int i = 1; i <= ncnt; i ++){
        for(int j = 1; j <= m && (ll)prime[i] * prime[i] <= a[j]; j ++){
            g[j] = (g[j] +P- (ll)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]+P)%P)%P;
        }
    }
}

ll solve(ll x){
    if(x <= 1) return 0;
		memset(g,0,sizeof(g));
		memset(id1,0,sizeof(id1));
		memset(id2,0,sizeof(id2));
		memset(flag,0,sizeof(flag));
		memset(prime,0,sizeof(prime));
		memset(sum,0,sizeof(sum));
		memset(a,0,sizeof(a));
		m=ncnt=0;
    return n = x, init(n), g[ID(n)]%P;
}
ll nn,p[N],s[N],ans;
int main(){
	cin>>nn;
	solve(nn);
    //puts("!");
	for(int i=1;i<=ncnt;i++){
		//cout<<"pri="<<prime[i]<<endl;
		int t=0;
		p[0]=1;
		while(p[t]<=nn/prime[i]) t++,p[t]=p[t-1]*prime[i];
        //cout<<"t="<<t<<endl;
		for(int i=0;i<=t;i++) s[i]=nn/p[i]%P;
		for(int i=0;i<t;i++) (s[i]+=P-s[i+1])%=P;//cout<<s[i]<<" "; puts("");
		for(int y=2;y<=t;y++){
			for(int x=0;x<y;x++){
				(ans+=1ll*(prime[i]%P)*(y-max(x,1))%P*s[x]%P*s[y]%P)%=P;
			}
		}
	}
	//cout<<"ans="<<ans<<endl;
	for(ll i=1;i<=nn;){
		ll x=nn/i,j=nn/x;
		//cout<<i<<" "<<j<<" "<<x<<endl;
		(ans+=1ll*((g[ID(j)]%P)+P-(g[ID(i-1)]%P))*(x%P)%P*((nn-x)%P)%P)%=P;
		
		i=j+1;
		//cout<<"ans="<<ans<<endl;
	}
	ans=(ans%P+P)%P;
	cout<<(ans+ans)%P<<endl;
}

PN筛

题目链接

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+5;
const ll P=(1ll<<57)*29+1,V2=(P+1)/2;
ll mul(ll a,ll x){
	return ((__int128)a)*x%P;
}
ll fpw(ll a,ll x){
    ll s=1;
    for(;x;x>>=1,a=mul(a,a)) if(x&1) s=mul(s,a);
    return s;
}
void mod(ll& x){
    x=(x%P+P)%P;
}
ll sum(ll n){
    return mul(mul(n,n+1),V2);
}
ll iv[N],sv[N],pr[N],pt,ans,n;
bool vs[N];
void dfs(int k,ll c,ll s){
    if(k>pt || pr[k]>n/c/pr[k]){
        (ans+=mul(c,mul(s,sum(n/c))))%=P;
        return;
    }
    ll t=c;
    dfs(k+1,t,s);
    for(int i=1;;i++){
        if(pr[k]>n/t) break;
        t*=pr[k];
        if(i>=2) dfs(k+1,t,P-mul(s,sv[i]));
    }
}
void work(){
    cin>>n;
    ans=0;
    dfs(1,1,1);
    cout<<mul(ans,fpw(n,P-2))<<endl;
}
int main()
{
    int n=1e6;
    iv[1]=1;
    for(int i=2;i<=n;++i){
          iv[i]=mul((P-P/i),iv[P%i]);
          sv[i]=mul(iv[i],iv[i-1]);
    }
    for(int i=2;i<=n;i++){
        if(!vs[i]) pr[++pt]=i;
        for(int j=1;j<=pt && pr[j]<=n/i;j++){
            int x=i*pr[j];
            vs[x]=1;
            if(!(i%pr[j])){
                break;
            }
        }
    }
    int T; cin>>T; while(T--) work();
    return 0;
}