2705: [SDOI2012]Longge的问题

题目链接

题目大意：求∑gcd(i, N)(1<=i <=N)

题解：

∑i=1ngcd(i,n)

=∑d|nd∑i=1n[gcd(i,n)==d]

令n′=[n/d]

=∑d|nd∑i=1n′[gcd(i,n′)==1]

=∑d|nd∗ϕ(n′)

因为范围太大没法筛，直接求就可以了

我的收获：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define ll long long ll n,ans;ll phi(ll x){    ll ret=x;    for(ll i=2;i*i<=x;i++)    if(x%i==0){ret-=ret/i;while(x%i==0) x/=i;}    if(x>1) ret-=ret/x;    return ret;}void work(){    cin>>n;    for(ll i=1;i*i<=n;i++)    if(n%i==0){        ans+=i*phi(n/i);        if(i*i<n) ans+=(n/i)*phi(i);    }    cout<<ans<<endl;} int main(){    work();    return 0;}