huntian oy

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{i}gcd(i^a?j^a,i^b?j^b)(gcd(i,j)==1) \\ =\sum _{i=1}^n\sum _{j=1}^i(i?j)(gcd(i,j)==1) \\ =\sum _{i=1}^{n}i\sum _{j=1}^i(gcd(i,j)==1)?\sum _{i=1}^n\sum _{j=1}^{i}j(gcd(i,j)==1) \\ =\sum _{i=1}^{n}i?(i)?\sum _{i=1}^n\frac{i?(i)+(i==1)}{2} \\ =\sum _{i=1}^n\frac{i?(i)?(i==1)}{2} \\ 然后套路地變成求S(n)=\sum _{i=1}^{n}i?(i) \\ g(n)=n?(n) \\ 這里直接套用杜教篩化簡(jiǎn)得到g(1)S(n)=\sum _{i=1}^n(f?g)(i)?\sum _{i=2}f(i)S(\frac{n}{i}) \\ (f?g)(i)=\sum _{d\mid n}f(d)?g(\frac{n}{d})=\sum _{d\mid n}f(d)?\frac{n}{d}?(\frac{n}{d}) \\ 另f(d)=d，則有S(n)=\sum _{i=1}^n{i}^2?\sum _{i=2}iS(\frac{n}{i}) \\ 然后直接套杜教篩即可。 \end{gathered}$$

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, inv2 = 500000004, inv6 = 166666668;int prime[N], cnt;ll phi[N];bool st[N];ll quick_pow(ll a, ll n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;n >>= 1;a = a * a % mod;}return ans; }void init() {phi[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;phi[i] = i - 1;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];break;}phi[i * prime[j]] = phi[i] * (prime[j] - 1);}}for(int i = 1; i < N; i++) {phi[i] = (1ll * phi[i] * i + phi[i - 1]) % mod;} }ll calc(int n) {return 1ll * n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod; }unordered_map<int, ll> ans_s;ll S(int n) {if(n < N) return phi[n];if(ans_s.count(n)) return ans_s[n];ll ans = calc(n);for(int l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans - 1ll * (r + l) * (r - l + 1) / 2 % mod * S(n / l) % 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();int T = read();while(T--) {int n = read(), a = read(), b = read();printf("%lld\n", (S(n) - 1 + mod) % mod * inv2 % mod);}return 0; } 創(chuàng)作挑戰(zhàn)賽新人創(chuàng)作獎(jiǎng)勵(lì)來(lái)咯，堅(jiān)持創(chuàng)作打卡瓜分現(xiàn)金大獎(jiǎng)

總結(jié)

以上是生活随笔為你收集整理的HDU 6706 huntian oy （欧拉函数 + 杜教筛）的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺(jué)得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。