qq_41997978的博客

∑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 =...

∑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⌋......

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...

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...

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[...