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求 ∑ i = 1 n f ( i ) , f ( p k ) = p k × ( p k − 1 ) \sum \limits_{i = 1} ^{n} f(i), f(p ^ k) = p ^ k \times(p ^ k - 1) i=1∑nf(i),f(pk)=pk×(pk−1)，最后对 m o d 1 e 9 + 7 \bmod 1e9 + 7 mod1e9+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 | 1
using 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 = 1 n i [ i ∈ p r i m e s ] \sum\limits_{i = 1} ^{n} i[i \in primes] i=1∑ni[i∈primes]
即是求 f ( p ) = p f(p) = p f(p)=p，直接求它的 g g 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 | 1
using 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
f ( 1 ) = 1 f ( p c ) = p ⊕ c f ( a b ) = f ( a ) ( b ) a , b 互 质 求 ∑ i = 1 n f ( i ) f(1) = 1\\ f(p ^ c) = p \oplus c\\ f(ab) = f(a)(b)\ a, b互质\\ 求\sum_{i = 1} ^{n} f(i)\\ f(1)=1f(pc)=p⊕cf(ab)=f(a)(b) a,b互质求i=1∑nf(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 | 1
using 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 | 1
using 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;
}
