Front cheese
- Number theory block, a little inversion Mobius
text
Questions asked
\ [\ sum_ {i = 1 } ^ {n} \ sum_ {j = 1} ^ {n} gcd (i, j) \]
Then go heavy.
It is contemplated that the enumeration \ (\ n) within \ (the p-\) , the contribution of the answers into \ (p \ times gcd (i , j) = p \) the number of the number of
\[\sum_{p=1}^{n}p\sum_{i=1}^{n}\sum_{j=1}^{n}[gcd(i,j)=p]\]
Simplification formula
\[\sum_{p=1}^{n}p\sum_{i=1}^{\left \lfloor \frac{n}{p} \right \rfloor }\sum_{j=1}^{\left \lfloor \frac{n}{p} \right \rfloor }[gcd(i,j)=1]\]
Since \ ([gcd (i, j ) = 1] \) only \ (gcd (i, j) = 1 \) have \ (1 \) contributions or not, it can be replaced with \ (\ varepsilon (gcd (i, j)) \)
\ (\ varepsilon (x) \ ) is, when \ (x = 1 \) when answers contributions \ (1 \) , otherwise \ (0 \)
So now converted to
\[\sum_{p=1}^{n}p\sum_{i=1}^{\left \lfloor \frac{n}{p} \right \rfloor }\sum_{j=1}^{\left \lfloor \frac{n}{p} \right \rfloor }\varepsilon(gcd(i,j))\]
By the \ (Dirichlet \) was convolution
\(\varepsilon =\mu*1 \Leftrightarrow \varepsilon(n)=\sum_{d|n}\mu(d)\)
So the original formula is converted to
\ [\ sum_ {p = 1 } ^ {n} p \ sum_ {i = 1} ^ {\ left \ lfloor \ frac {n} {p} \ right \ rfloor} \ sum_ {j = 1} ^ {\ left \ lfloor \ frac {n} {p} \ right \ rfloor} \ sum_ {d | gcd (i, j)} \ mu (d) \]
Transforming sums order
\[\sum_{p=1}^{n}p\sum_{d=1}{\mu(d)}\sum_{i=1}^{\left \lfloor \frac{n}{p} \right \rfloor }[d\ |\ i]\sum_{j=1}^{\left \lfloor \frac{n}{p} \right \rfloor}[d\ | \ j]\]
The latter two requirements is the \ (\ left \ lfloor \ frac {n} {p} \ right \ rfloor \) in \ (D \) multiple, easily know the answer is \ (\ left \ lfloor \ frac {n } {pd} \ right \ rfloor \)
So the answer into
\[ans=\sum_{p=1}^{n}p\sum_{d=1}\mu(d)\left \lfloor \frac{n}{pd} \right \rfloor\left \lfloor \frac{n}{pd} \right \rfloor\]
And that thing back block and can use the prefix number theory.
\(Code：\)
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define int long long
inline int read(){
register int x=0,f=0,ch=getchar();
while('0'>ch||ch>'9')f^=ch=='-',ch=getchar();
while('0'<=ch&&ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=getchar();
return f?-x:x;
}
const int MAX=100005;
int n,tot,p[MAX],f[MAX],mu[MAX];
inline void Gmu(){
mu[1]=1;tot=0;
for(register int i=2;i<=n;++i){
if(!f[i]){
p[++tot]=i;
mu[i]=-1;
}
for(register int j=1;j<=tot&&i*p[j]<=n;++j){
f[i*p[j]]=1;
if(i%p[j]==0){
mu[i*p[j]]=0;
break;
}
mu[i*p[j]]=-mu[i];
}
}
for(register int i=1;i<=n;++i)mu[i]+=mu[i-1];
}
int res,ans;
inline void solve(int n,int p){
res=0;
for(register int l=1,r;l<=n;l=r+1){
r=n/(n/l);
res+=(mu[r]-mu[l-1])*(n/l)*(n/l);
}
ans+=res*p;
}
signed main(){
n=read();
Gmu();
for(register int p=1;p<=n;++p)solve(n/p,p);
printf("%lld\n",ans);
return 0;
}What's written questions trouble private letter, thank you