与Min_25筛有关的一些模板

模板

求 $∑i=1nf(i),f(pk)=pk×(pk−1)\sum \limits_{i = 1} ^{n} f(i), f(p ^ k) = p ^ k \times(p ^ k - 1)$ ，最后对 $mod1e9+7\bmod 1e9 + 7$ ，这个函数是个积性函数。

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e6 + 10, mod = 1e9 + 7, inv2 = 500000004, inv6 = 166666668;namespace MIN_25 {int prime[N], id1[N], id2[N], cnt, m, T;ll g1[N], g2[N], sum1[N], sum2[N], a[N], n;bool st[N];int ID(ll x) {return x <= T ? id1[x] : id2[n / x];}ll calc2(ll x) {x %= mod;return (x * (x + 1) % mod * inv2 % mod - 1 + mod) % mod;}ll calc1(ll x) {x %= mod;return (x * (x + 1) % mod * (2 * x + 1) % mod * inv6 % mod - 1 + mod) % mod;}ll f(ll x) {x %= mod;return x * (x - 1) % mod;}void init() {T = sqrt(n + 0.5);for(int i = 2; i <= T; i++) {if(!st[i]) {prime[++cnt] = i;sum1[cnt] = (sum1[cnt - 1] + 1ll * i * i) % mod;sum2[cnt] = (sum2[cnt - 1] + i) % mod;}for(int j = 1; j <= cnt && 1ll * i * prime[j] <= T; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {break;}}}for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);a[++m] = n / l;if(a[m] <= T) id1[a[m]] = m;else id2[n / a[m]] = m;g1[m] = calc1(a[m]);g2[m] = calc2(a[m]);}for(int j = 1; j <= cnt; j++) {for(int i = 1; i <= m && 1ll * prime[j] * prime[j] <= a[i]; i++) {g1[i] = ((g1[i] - 1ll * prime[j] * prime[j] % mod * (g1[ID(a[i] / prime[j])] - sum1[j - 1]) % mod) % mod + mod) % mod;g2[i] = ((g2[i] - 1ll * prime[j] * (g2[ID(a[i] / prime[j])] - sum2[j - 1]) % mod) % mod + mod) % mod;}}}ll solve(ll n, int m) {if(n < prime[m]) return 0;ll ans = ((g1[ID(n)] - sum1[m - 1] - g2[ID(n)] + sum2[m - 1] % mod) + mod) % mod;for(int j = m; j <= cnt && prime[j] * prime[j] <= n; j++) {for(ll i = prime[j]; i * prime[j] <= n; i *= prime[j]) {ans = (ans + f(i) * solve(n / i, j + 1) % mod + f(i * prime[j])) % mod;}}return ans;}ll solve(ll x) {n = x;init();return solve(n, 1) + 1;}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cout << MIN_25::solve(read()) << endl;return 0;
}

求 $∑i=1ni[i∈primes]\sum\limits_{i = 1} ^{n} i[i \in primes]$

即是求 $f (p) = p$ ，直接求它的 $g$ 函数即可。

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e6 + 10, mod = 1e9 + 7, inv2 = 500000004, inv6 = 166666668;int prime[N], id1[N], id2[N], cnt, m, T;ll g[N], sum[N], a[N], n;bool st[N];int ID(ll x) {return x <= T ? id1[x] : id2[n / x];
}ll calc(ll x) {return x * (x + 1) / 2 - 1;
}void init() {T = sqrt(n + 0.5);for(int i = 2; i <= T; i++) {if(!st[i]) {prime[++cnt] = i;sum[cnt] = sum[cnt - 1] + i;}for(int j = 1; j <= cnt && 1ll * i * prime[j] <= T; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {break;}}}for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);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 j = 1; j <= cnt; j++) {for(int i = 1; i <= m && 1ll * prime[j] * prime[j] <= a[i]; i++) {g[i] = g[i] - 1ll * prime[j] * (g[ID(a[i] / prime[j])] - sum[j - 1]);}}
}ll solve(ll x) {n = x;init();return g[ID(x)];
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cout << solve(read()) << endl;return 0;
}

#6053. 简单的函数

$1\\ f(p ^ c) = p \oplus c\\ f(ab) = f(a)(b)\ a, b互质\\ 求\sum_{i = 1} ^{n} f(i)\\$

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e6 + 10, mod = 1e9 + 7, inv2 = 500000004, inv6 = 166666668;namespace MIN_25 {int prime[N], id1[N], id2[N], cnt, m, T;ll g1[N], g2[N], sum1[N], sum2[N], a[N], n;bool st[N];int ID(ll x) {return x <= T ? id1[x] : id2[n / x];}ll calc(ll x) {x %= mod;return (x * (x + 1) % mod * inv2 % mod - 1 + mod) % mod;}void init() {T = sqrt(n + 0.5);for(int i = 2; i <= T; i++) {if(!st[i]) {prime[++cnt] = i;sum1[cnt] = (sum1[cnt - 1] + i) % mod;sum2[cnt] = (sum2[cnt - 1] + 1) % mod;}for(int j = 1; j <= cnt && 1ll * i * prime[j] <= T; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {break;}}}for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);a[++m] = n / l;if(a[m] <= T) id1[a[m]] = m;else id2[n / a[m]] = m;g1[m] = calc(a[m]);g2[m] = (a[m] - 1) % mod;}for(int j = 1; j <= cnt; j++) {for(int i = 1; i <= m && 1ll * prime[j] * prime[j] <= a[i]; i++) {g1[i] = ((g1[i] - 1ll * prime[j] * (g1[ID(a[i] / prime[j])] - sum1[j - 1]) % mod) % mod + mod) % mod;g2[i] = ((g2[i] - 1ll * (g2[ID(a[i] / prime[j])] - sum2[j - 1]) % mod) % mod + mod) % mod;}}}ll solve(ll n, int m) {if(n < prime[m]) return 0;ll ans = ((g1[ID(n)] - sum1[m - 1] - g2[ID(n)] + sum2[m - 1]) % mod + mod) % mod;for(int j = m; j <= cnt && 1ll * prime[j] * prime[j] <= n; j++) {for(ll i = prime[j], c = 1; i * prime[j] <= n; i *= prime[j], c++) {ans = (ans + 1ll * (prime[j] ^ c) * solve(n / i, j + 1) % mod + 1ll * (prime[j] ^ (c + 1))) % mod;}}return ans;}ll solve(ll x) {n = x;init();return (solve(x, 1) + 1 + (x > 1) * 2) % mod;}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cout << MIN_25::solve(read()) << endl;return 0;
}

#6235. 区间素数个数

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e6 + 10, mod = 1e9 + 7, inv2 = 500000004, inv6 = 166666668;namespace MIN_25 {int prime[N], id1[N], id2[N], m, cnt, T;ll a[N], g[N], sum[N], n;bool st[N];int ID(ll x) {return x <= T ? id1[x] : id2[n / x];}void init() {T = sqrt(n + 0.5);for(int i = 2; i <= T; i++) {if(!st[i]) {prime[++cnt] = i;sum[cnt] = sum[cnt - 1] + 1;}for(int j = 1; j <= cnt && 1ll * i * prime[j] <= T; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {break;}}}for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);a[++m] = n / l;if(a[m] <= T) id1[a[m]] = m;else id2[n / a[m]] = m;g[m] = a[m] - 1;}for(int j = 1; j <= cnt; j++) {for(int i = 1; i <= m && 1ll * prime[j] * prime[j] <= a[i]; i++) {g[i] = g[i] - 1ll * (g[ID(a[i] / prime[j])] - sum[j - 1]);}}}ll solve(ll x) {n = x;init();return g[ID(n)];}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);cout << MIN_25::solve(read()) << endl;return 0;
}