[CQOI2015]选数（杜教筛）
[CQOI2015]選數
推式子
根據題意可寫出式子：
∑a1=LH∑a2=LH?∑an=LH[gcd(a1,a2…an)=k]∑a1=?Lk??Hk?∑a2=?Lk??Hk??∑an=?Lk??Hk?[gcd(a1,a2…an)=k]∑k=1?Hk?μ(k)(?Hkd???Lkd?+1)n提前處理一下左右端點∑k=1?Hk?μ(k)(?Hkd???L?1kd?)n\sum_{a_1 = L} ^{H} \sum_{a_2 = L} ^{H} \dots \sum_{a_n = L} ^{H}[gcd(a_1, a_2 \dots a_n) = k]\\ \sum_{a_1 = \lceil \frac{L}{k} \rceil} ^{\lfloor \frac{H}{k} \rfloor} \sum_{a_2 = \lceil \frac{L}{k} \rceil} ^{\lfloor \frac{H}{k} \rfloor}\dots \sum_{a_n = \lceil \frac{L}{k} \rceil} ^{\lfloor \frac{H}{k} \rfloor}[gcd(a_1, a_2 \dots a_n) = k]\\ \sum_{k = 1} ^{\lfloor \frac{H}{k} \rfloor} \mu(k) \left( \lfloor \frac{H}{kd} \rfloor - \lceil \frac{L}{kd} \rceil + 1 \right) ^n\\ 提前處理一下左右端點\\ \sum_{k = 1} ^{\lfloor \frac{H}{k} \rfloor} \mu(k) \left( \lfloor \frac{H}{kd} \rfloor - \lfloor \frac{L - 1}{kd} \rfloor\right) ^n\\ a1?=L∑H?a2?=L∑H??an?=L∑H?[gcd(a1?,a2?…an?)=k]a1?=?kL??∑?kH???a2?=?kL??∑?kH????an?=?kL??∑?kH???[gcd(a1?,a2?…an?)=k]k=1∑?kH???μ(k)(?kdH????kdL??+1)n提前處理一下左右端點k=1∑?kH???μ(k)(?kdH????kdL?1??)n
非常簡單的反演公式，所以我們只要按照這個公式套上杜教篩就行了。
代碼
/*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;int prime[N], mu[N], cnt; ll n, k, L, H;bool st[N];void init() {mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;mu[i] = -1;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {mu[i] += mu[i - 1];} }ll quick_pow(ll a, ll n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans; }map<ll, ll> ans_s;ll S(ll n) {if(n < N) return mu[n];if(ans_s.count(n)) return ans_s[n];ll ans = 1;for(ll l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = ((ans - (r - l + 1) * S(n / l) % mod) % mod + mod) % mod;}return ans_s[n] = ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();n = read(), k = read(), L = read(), H = read();ll ans = 0;L = (L - 1) / k, H = H / k;for(ll l = 1, r; l <= H; l = r + 1) {r = min( L / l ? L / (L / l) : inf , H / (H / l));ans = (ans + 1ll * (((S(r) - S(l - 1)) % mod + mod) % mod) * quick_pow((H / l) - (L / l), n) % mod) % mod;}cout << ans << endl;return 0; }總結
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