【BZOJ 4176】 Lucas的数论 - 杜教筛

　　题意：求∑ni=1∑nj=1σ0(ij)
　　
　　把约数个数给展开来写，然后注意到d∣ij等价于dgcd(i,d)∣j，于是代入，然后可以拿个求和指标替换掉gcd(i,d)，再然后可以用个μ展开一波，最后会发现有几个相同的求和指标，等价于求一个和然后平方，于是可以得到最后的式子是
　　

∑k=1nμ(k)⎛⎝⎜∑p=1⌊nk⌋⌊nkp⌋⎞⎠⎟2

　　设Sr(n)=∑rk=1⌊nk⌋，我们知道有⌊⌊nk⌋p⌋=⌊nkp⌋，那么式子可以变为
　　

∑k=1nμ(k)S⌊nk⌋(n)2

　　记Fn=∑ni=1μ(i)，则有Fn=1−∑nk=2F(⌊nk⌋)，那么对于每一段相同的n/k，两个东西都可以直接计算。所以这个的复杂度相当于是一次杜教筛。

懒癌晚期不想写公式结果还是写了那么多(?)

/*    I will chase the meteor for you, a thousand times over.    Please wait for me, until I fade forever.    Just 'coz GEOTCBRL.*/#include <bits/stdc++.h>using namespace std;#define fore(i,u)  for (int i = head[u] ; i ; i = nxt[i])#define rep(i,a,b) for (int i = a , _ = b ; i <= _ ; i ++)#define per(i,a,b) for (int i = a , _ = b ; i >= _ ; i --)#define For(i,a,b) for (int i = a , _ = b ; i <  _ ; i ++)#define Dwn(i,a,b) for (int i = ((int) a) - 1 , _ = (b) ; i >= _ ; i --)#define fors(s0,s) for (int s0 = (s) , _S = s ; s0 ; s0 = (s0 - 1) & _S)#define foreach(it,s) for (__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)#define mp make_pair#define pb push_back#define pii pair<int , int>#define fir first#define sec second#define MS(x,a) memset(x , (a) , sizeof (x))#define gprintf(...) fprintf(stderr , __VA_ARGS__)#define gout(x) std::cerr << #x << "=" << x << std::endl#define gout1(a,i) std::cerr << #a << '[' << i << "]=" << a[i] << std::endl#define gout2(a,i,j) std::cerr << #a << '[' << i << "][" << j << "]=" << a[i][j] << std::endl#define garr(a,l,r,tp) rep (__it , l , r) gprintf(tp"%c" , a[__it] , " \n"[__it == _])template <class T> inline void upmax(T&a , T b) { if (a < b) a = b ; }template <class T> inline void upmin(T&a , T b) { if (a > b) a = b ; }typedef long long ll;const int maxn = 8000007;const int maxm = 200007;const int mod = 1000000007;const int inf = 0x7fffffff;const double eps = 1e-7;typedef int arr[maxn];typedef int adj[maxm];inline int fcmp(double a , double b) {    if (fabs(a - b) <= eps) return 0;    if (a < b - eps) return -1;    return 1;}inline int add(int a , int b) { return ((ll) a + b) % mod ; }inline int mul(int a , int b) { return ((ll) a * b) % mod ; }inline int dec(int a , int b) { return add(a , mod - b % mod) ; }inline int Pow(int a , int b) {    int t = 1;    while (b) {        if (b & 1) t = mul(t , a);        a = mul(a , a) , b >>= 1;    }    return t;}#define gc getchar#define idg isdigit#define rd RD<int>#define rdll RD<long long>template <typename Type>inline Type RD() {    char c = getchar(); Type x = 0 , flag = 1;    while (!idg(c) && c != '-') c = getchar();    if (c == '-') flag = -1; else x = c - '0';    while (idg(c = gc()))x = x * 10 + c - '0';    return x * flag;}inline char rdch() {    char c = gc();    while (!isalpha(c)) c = gc();    return c;}#undef idg#undef gc// beginningarr mu , vis , pr;int tot , N , n;inline void sieve(int n) {    mu[1] = 1;    rep (i , 2 , n) {        if (!vis[i])            pr[++ tot] = i , mu[i] = -1;        rep (j , 1 , tot) if (i * pr[j] > n) break; else {            vis[i * pr[j]] = 1;            if (i % pr[j] == 0) break;            mu[i * pr[j]] = -mu[i];        }    }    rep (i , 1 , n) mu[i] = (mu[i - 1] + mu[i]) % mod;}void input() {    n = rd() , N = int(pow(n , 0.75) + 0.5);    sieve(N);}const int cut = 10007;map<int , int> ms[cut];int calc(int n) {    if (n <= N)        return mu[n];    if (ms[n % cut].count(n)) return ms[n % cut][n];    int ret = 1;    for (int l = 2 , r = 0; l <= n; l = r + 1) {        r = n / (n / l);        int tmp = calc(n / l);        ret = dec(ret , mul(r - l + 1 , tmp));    }    ms[n % cut][n] = ret;    return ret;}inline int get_sum(int n) {    int ret = 0;    for (int l = 1 , r = 0; l <= n; l = r + 1) {        r = n / (n / l);        ret = add(ret , mul(n / l , r - l + 1));    }    return mul(ret , ret);}void solve() {    int ans = 0;    for (int l = 1 , r = 0; l <= n; l = r + 1) {        r = n / (n / l);        int tmp = get_sum(n / l);        int res = dec(calc(r) , calc(l - 1));//      gprintf("[%d , %d]\n" , l , r);//      gout(tmp) , gout(res);        ans = add(ans , mul(tmp , res));    }    if (ans < 0) ans += mod;    printf("%d\n" , ans);}int main() {    #ifndef ONLINE_JUDGE        freopen("data.txt" , "r" , stdin);    #endif    input();    solve();    return 0;}