bzoj 2705: [SDOI2012]Longge的问题 歐拉函數

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<string>
#include<queue>
using namespace std;
#ifdef WIN32
#define LL "%I64d"
#else
#define LL "%lld"
#endif
#define MAXN 1100000
#define MAXV MAXN*2
#define MAXE MAXV*2
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
typedef long long qword;
inline int nextInt()
{
        char ch;
        int x=0;
        bool flag=false;
        do
                ch=getchar(),flag=(ch==‘-‘)?true:flag;
        while(ch<‘0‘||ch>‘9‘);
        do x=x*10+ch-‘0‘;
        while (ch=getchar(),ch<=‘9‘ && ch>=‘0‘);
        return x*(flag?-1:1);
}

qword n,m;
qword prime[MAXN],topp=-1;
bool pflag[MAXN];
int gcd(int x,int y)
{
        return (x%y==0) ? y : gcd(y,x%y);
}
void init()
{
        int i,j;
        for (i=2;i<MAXN;i++)
        {
                if (!pflag[i])
                {
                        prime[++topp]=i;
                }
                for (j=0;j<=topp && i*prime[j]<MAXN;j++)
                {
                        pflag[i*prime[j]]=true;
                        if (i%prime[j]==0)
                        {
                                break;
                        }
                }
        }
}
qword phi(qword x)
{
        int i;
        qword ret=1;
        for (i=0;i<=topp && prime[i]*prime[i]<=x;i++)
        {
                if(x%prime[i]==0)
                {
                        ret*=prime[i]-1;
                        x/=prime[i];
                        while (x%prime[i]==0)
                        {
                                ret*=prime[i];
                                x/=prime[i];
                        }
                }
        }
        if (x>1)
        {
                ret*=x-1;
        }
        return ret;
}
int main()
{
        freopen("input.txt","r",stdin);
        //freopen("output.txt","w",stdout);
        qword i;
        scanf("%d",&n);
        qword ans1=0,ans2=0;
    //    for (i=1;i<=n;i++)
    //            ans1+=gcd(i,n);
    //    printf("%d\n",ans1);
        init();
        qword l=ceil(sqrt(n));
        for (i=1;i*i<n;i++)
        {
                if (n%i==0)
                {
                        ans2+=(qword)i*phi(n/i);
                        ans2+=(qword)(n/i)*phi(i);
                }
        }
        if (l*l==n)
        {
                ans2+=(qword)(n/l)*phi(l);
        }
        printf(LL"\n",ans2);
        return 0;
}