BZOJ 2705: [SDOI2012]Longge的问题 | 数论

题目:
http://www.lydsy.com/JudgeOnline/problem.php?id=2705

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题解:

求Σgcd(i,n)

因为gcd(i,n)一定是n的约数

所以我们枚举n的约数,对于每个约数d,他的贡献是d*φ(n/d)

所以暴力枚举约数,然后暴力求欧拉即可

#include<cstdio>
typedef long long ll;
using namespace std;
ll n,ans;
ll oula(ll n)
{
    ll ans=n,a=n;
    for(ll i=2;i*i<=n;i++)
    if(a%i==0)
    {
        ans-=ans/i;
        while(a%i==0)
        a/=i;
    }
    if(a>1) ans-=ans/a;
    return ans;
}
int main()
{
    scanf("%lld",&n);
    for (int i=1;i*i<=n;i++)
    if (n%i==0)
    {
        ans+=oula(n/i)*i;
        if (i*i!=n) ans+=oula(i)*n/i;
    }
    printf("%lld\n",ans);
    return 0;
}

 

转载于:https://www.cnblogs.com/mrsheep/p/8192796.html