qq_41997978的博客

题目大意：让你求 ∑i=1n∑j=1m[gcd(i,j)=p],p∈prime\sum_{i = 1}^n\sum_{j = 1}^m[gcd(i,j) = p],p \in prime∑i=1n​∑j=1m​[gcd(i,j)=p],p∈prime转化一下式子，枚举 p：∑p=2n∑i=1n∑j=1m[gcd(i,j)=p],p∈prime\sum_{p = 2}^n\sum_{i = 1...

如图：求 ∏i=1n∏j=1n∏k=1nmgcd(i,j)[k∣gcd(i,j)]\prod_{i = 1}^n\prod_{j = 1}^n\prod_{k = 1}^nm^{gcd(i,j)[k | gcd(i,j)]}∏i=1n​∏j=1n​∏k=1n​mgcd(i,j)[k∣gcd(i,j)]把求和符号放到幂次就变成求和符号，幂次式子变成：∑i=1n∑j=1n∑k=1ngcd(i,j...

题目大意：让你求：∑i=1n∑j=1n∑k=1nϕ(i)ϕ(j2)ϕ(k3)ϕ(i)ϕ(j)ϕ(i)ϕ(gcd(i,j,k))让你求：\sum_{i = 1}^n\sum_{j = 1}^n\sum_{k = 1}^n\frac{\phi(i)\phi(j ^ 2)\phi(k^3)}{\phi(i)\phi(j)\phi(i)}\phi(gcd(i,j,k))让你求：i=1∑n​j=1∑n​...

化简 fn(k)=∑l1=1n∑l2=1n...∑lk=1n(gcd(l1,l2,...,lk))2f_n(k)=\sum_{l_1 = 1}^n\sum_{l_2 = 1}^n...\sum_{l_k = 1}^n(gcd(l_1,l_2,...,l_k))^2fn​(k)=l1​=1∑n​l2​=1∑n​...lk​=1∑n​(gcd(l1​,l2​,...,lk​))2=∑d=1n∑l1...

∑i1=1h∑i2=1h∑i3=1h...[gcd(i1,i2,i3,..)=k]\sum_{i_1 = 1}^h\sum_{i_2 = 1}^h\sum_{i_3 = 1}^h...[gcd(i_1,i_2,i_3,..) = k]i1​=1∑h​i2​=1∑h​i3​=1∑h​...[gcd(i1​,i2​,i3​,..)=k]=∑i1=1⌊hk⌋∑i2=1⌊hk⌋∑i3=1⌊hk⌋......

题目大意：让你求：∑i=1ngcd(⌊n3⌋,i)mod  998322353\sum_{i = 1}^ngcd(\lfloor\sqrt[3]{n}\rfloor,i) \mod 998322353∑i=1n​gcd(⌊3n​⌋,i)mod998322353n≤1021n \leq 10 ^ {21}n≤1021题解：明显可以按n...

∑i=1n∑j=1ni∗j∗gcd(i,j)\sum_{i = 1}^n\sum_{j = 1}^ni * j * gcd(i,j)i=1∑n​j=1∑n​i∗j∗gcd(i,j)=∑d=1n∑i=1⌊nd⌋∑j=1⌊nd⌋i∗j∗d3∗[gcd(i,j)=1] =\sum_{d = 1}^{n}\sum_{i = 1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j =...

令 g[i]=∑d∣idg[i] = \sum_{d | i}dg[i]=∑d∣i​d，题目要求：∑i=1n∑j=1mg[gcd(i,j)]∗[g[gcd(i,j)]<=a]\sum_{i = 1}^n\sum_{j = 1}^mg[gcd(i,j)] * [g[gcd(i,j)] <= a]∑i=1n​∑j=1m​g[gcd(i,j)]∗[g[gcd(i,j)]&...

代码：#include<bits/stdc++.h>using namespace std;const int mod = 1e9 + 7;const int maxn = 2e6 + 10;typedef long long ll;ll fpow(ll a,ll b) { ll r = 1; while(b) { if(b & 1) r = r * ...

题解：假设n&lt;mn &lt; mn<m，一波化简：∑i=1n∑j=1mlcm(i,j)=∑i=1n∑j=1mi∗jgcd(i,j)=∑d=1n∑i=1n∑j=1mi∗jd∗[gcd(i,j)=d]=∑d=1n∑i=1⌊nd⌋∑j=1⌊md⌋i∗j∗d∗[gcd(i,j)=1]\sum_{i = 1}^n\sum_{j = 1}^{m}lcm(i,j) = \sum...

∑x=ab∑y=cd[gcd(x,y]=k]\sum_{x = a}^b\sum_{y = c}^d [gcd(x,y] = k]∑x=ab​∑y=cd​[gcd(x,y]=k]，下界不是从1开始好像有点麻烦？可以容斥一下：设solve(a,b)函数解决的是：∑x=1a∑y=1b[gcd(x,y]=k]\sum_{x = 1}^a\sum_{y = 1}^b [gcd(x,y] = k]∑x=...

首先 d(i∗j)=∑x∣i∑y∣j[gcd(x,y)=1]d(i * j) = \sum_{x | i}\sum_{y|j}[gcd(x,y) = 1]d(i∗j)=∑x∣i​∑y∣j​[gcd(x,y)=1]，证明感性理解一下就好。然后整个式子变成：∑i=1N∑j=1M∑x∣i∑y∣j[gcd(x,y)=1]\sum_{i = 1}^N\sum_{j = 1}^M\sum_{x | i}\...

代码：#include <bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 2e5 + 10;const ll mod = 1ll << 32;bool ispri[maxn];ll pri[maxn], num, tot;ll pw[maxn];ll g[...