Mobius inversion formula we have derived before for:
\(f(n)=\sum_{d|n}g(d),g(n)=\sum_{d|n}f(d)\times\mu(\frac n d)\)
Specific derivation can refer to my previous one blog .
And before we derive the Mobius inversion of this formula, he had received another equally important equation:
\(\sum_{d|n}\mu(d)=[n=1]\)
That is, \ (\ * 1 MU = \ Epsilon \) .
This equation dim it?
We convert it to a little bit, so that \ (the n-= gcd (i, J) \) , this is the case, the equation becomes:
\(\sum_{d|gcd(i,j)}\mu(d)=[gcd(i,j)=1]\)
And we can exchange (\ Sigma \) \ to achieve some similar location \ (gcd (i, j) = 1 \) to solve the equation.
Example 1:
求：\(\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=1]\) 。
This is a specific application of the formulas I said above it, with the above formula, this problem is very simple matter.
\(\ \ \ \ \sum_{i=1}^n\sum_{j=1}^{m}[gcd(i,j)=1]\)
\(=\sum_{i=1}^n\sum_{i=1}^m\sum_{d|gcd(i,j)}\mu(d)\)
\(=\sum_{d=1}^n\mu(d)\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac m d\rfloor}\)
\(=\sum_{d=1}^n\mu(d)\times\lfloor\frac n d\rfloor\times\lfloor\frac m d\rfloor\)
Then it may be (O (\ sqrt {n} ) \) \ solved in time.
Because Example 1, the code is posted, when the demonstration:
#include<bits/stdc++.h>
#define ll long long
const int N=10000000;
int mu[N+5],vis[N+5],sum[N+5],p[N+5];
ll ans;
int n,m,q;
using namespace std;
void sieve(){
mu[1]=vis[1]=1;
for(int i=2;i<=N;i++){
if (!vis[i]) p[++p[0]]=i,mu[i]=-1;
for (int j=1;j<=p[0]&&i*p[j]<=N;j++){
vis[i*p[j]]=1;
if (i%p[j]==0){
mu[i*p[j]]=0;
break;
}
mu[i*p[j]]=-mu[i];
}
}
for (int i=1;i<=N;i++)
mu[i]+=mu[i-1];
}
int main(){
sieve();
scanf("%d",&q);
while (q--){
scanf("%d%d",&n,&m);
if (n>m) swap(n,m);
ans=0;
for (int l=1,r;l<=n;l=r+1){
r=min(n/(n/l),m/(m/l));
ans+=1ll*(mu[r]-mu[l-1])*(n/l)*(m/l);
}
printf("%lld\n",ans);
}
return 0;
}Example 2:
求：\(\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=k]\) 。
In fact, here only need a simple transformation:
\(\ \ \ \ \sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=k]\)
\(=\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac m d\rfloor}[gcd(i,j)=1]\)
Then the same as above it.
Example 3:
求：\(\sum_{i=1}^n\sum_{j=1}^mi\times j\times[gcd(i,j)=k]\)
This and the above, it is no big difference, here \ (i, j \) is only a by-product, as long as we shift \ (\ Sigma \) when not forget to \ (i, j \) impact of items generated that is can.
\(\ \ \ \ \sum_{i=1}^n\sum_{j=1}^mi\times j\times[gcd(i,j)=k]\)
\(=\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac m d\rfloor}i\times j\times k^2\times[gcd(i,j)=1]\)
\(=k^2\times\sum_{d=1}^{\lfloor\frac n d\rfloor}\mu(d)\times\sum_{i=1}^{\lfloor\frac n d\rfloor}i\times\sum_{j=1}^{\lfloor\frac m d\rfloor}j\)
Decree \ (sum [n] = \ sum_ {i = 1} ^ ni \) .
\(=k^2\times\sum_{d=1}^{\lfloor\frac n d\rfloor}\mu(d)\times sum[\lfloor\frac n d\rfloor]\times sum[\lfloor\frac m d\rfloor]\)
\ (O (\ sqrt {n }) \) Solution to.
To so much of it, and the rest, there are many, after all, Mobius inversion profound, after the write.