最近在搞min_25筛，就写几道筛法题吧

模板题，直接套板子就好了

//waz#include 
     
      using namespace std;#define mp make_pair#define pb push_back#define fi first#define se second#define ALL(x) (x).begin(), (x).end()#define SZ(x) ((int)((x).size()))typedef pair
      
        PII;typedef vector
       
         VI;typedef long long int64;typedef unsigned int uint;typedef unsigned long long uint64;#define gi(x) ((x) = F())#define gii(x, y) (gi(x), gi(y))#define giii(x, y, z) (gii(x, y), gi(z))int F(){    char ch;    int x, a;    while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');    if (ch == '-') ch = getchar(), a = -1;    else a = 1;    x = ch - '0';    while (ch = getchar(), ch >= '0' && ch <= '9')        x = (x << 1) + (x << 3) + ch - '0';    return a * x;}const int mod = 1e9 + 7;int inc(int a, int b) { a += b; return a >= mod ? a - mod : a; }int dec(int a, int b) { a -= b; return a < 0 ? a + mod : a; }int fpow(int a, int x){    int ret = 1;    for (; x; x >>= 1)    {        if (x & 1) ret = 1LL * ret * a % mod;        a = 1LL * a * a % mod;    }    return ret;}const int N = 1e5 + 10;int64 n;int p[N], pt, sump[N], sumf[N];bool fg[N];int sqrtn;namespace work1{     int f1[N << 1], f2[N << 1];     int id1[N], id2[N], idt;    int64 num[N << 1];    int id(int64 x) { if (x <= sqrtn) return id1[x]; else return id2[n / x]; }    void solve()    {        idt = 0;        for (int64 i = 1, j; i <= n; i = j + 1)        {            int64 t = n / i;            j = n / t;            if (t <= sqrtn) id1[t] = ++idt;            else id2[n / t] = ++idt;            num[idt] = t;            int64 c = t % mod;            f1[idt] = (c * (c + 1) / 2 + mod - 1) % mod;            f2[idt] = c - 1;        }        for (int j = 1; j <= pt; ++j)        {            int64 end = 1LL * p[j] * p[j];            for (int k = 1; num[k] >= end; ++k)            {                int i = id(num[k] / p[j]);                f1[k] = dec(f1[k], 1LL * dec(f1[i], sump[j - 1]) * p[j] % mod);                f2[k] = dec(f2[k], 1LL * dec(f2[i], j - 1));            }        }    }    int query(int64 x) { return inc(dec(f1[id(x)], f2[id(x)]), 2 * (x >= 2)); }}namespace work2{    int g[N << 1];    int id1[N], id2[N], idt;    int64 num[N << 1];    int id(int64 x) { if (x <= sqrtn) return id1[x]; else return id2[n / x]; }    int solve()    {        for (int64 i = 1, j; i <= n; i = j + 1)        {            int64 t = n / i;            j = n / t;            if (t <= sqrtn) id1[t] = ++idt;            else id2[n / t] = ++idt;            num[idt] = t;            g[idt] = work1::query(t);        }        for (int j = pt; j; --j)        {            for (int k = 1; num[k] >= 1LL * p[j] * p[j]; ++k)            {                int i = 1;                for (int64 m = p[j]; m * p[j] <= num[k]; ++i, m *= p[j])                {                    g[k] = inc(g[k], 1LL * dec(g[id(num[k] / m)], sumf[j]) * (p[j] ^ i) % mod);                    g[k] = inc(g[k], p[j] ^ (i + 1));                }            }        }        return inc(g[id(n)], 1);    }}int main(){    cin >> n;    sqrtn = sqrt(n);    for (int i = 2; i <= sqrtn; ++i)    {        if (!fg[i])        {            p[++pt] = i;        }        for (int j = 1; j <= pt && p[j] * i <= sqrtn; ++j)        {            fg[p[j] * i] = 1;            if (i % p[j] == 0) break;        }    }    for (int i = 1; i <= pt; ++i) sump[i] = inc(sump[i - 1], p[i]), sumf[i] = inc(sumf[i - 1], p[i] ^ 1);    work1::solve();    //cerr << work1::query(n) << endl;    cout << work2::solve() << endl;}