模板

求$\sum _{i=1}^{n}f(i),f(p^k)=p^k\times (p^k?1)$，最后對(duì)$\mathrm{m}\mathrm{o}\mathrm{d}1e9+7$，這個(gè)函數(shù)是個(gè)積性函數(shù)。

/*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$\sum _{i=1}^{n}i[i\in primes]$

即是求$f(p)=p$，直接求它的$g$函數(shù)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. 簡(jiǎn)單的函數(shù)

$$\begin{gathered} f(1)=1 \\ f(p^c)=p\oplus c \\ f(ab)=f(a)(b)a,b互質(zhì) \\ 求\sum _{i=1}^{n}f(i) \end{gathered}$$

/*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. 區(qū)間素?cái)?shù)個(gè)數(shù)

/*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; }

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