Longge's problem poj2480 欧拉函数，gcd

Longge's problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6918		Accepted: 2234

Description

Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N. 

"Oh, I know, I know!" Longge shouts! But do you know? Please solve it. 

Input

Input contain several test case. 
A number N per line. 

Output

For each N, output ,∑gcd(i, N) 1<=i <=N, a line

Sample Input

2
6

Sample Output

3
15

Source

 

 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <string.h>
#define MAXN 100010
#define INF 0x7fffffff
using namespace std;
long long a[110000]= {0},b[110000],nu,c[10000][2],cn;
void init()
{
    nu=0;
    long long i,j;
    for(i=2; i<110000; i++)
    {
        if(!a[i])
        {
            b[nu++]=i;
            j=i*i;
            while(j<110000)
            {
                a[j]=1;
                j+=i;
            }
        }
    }
}
void fun(long long x)
{
    long long i;
    memset(c,0,sizeof(c));
    cn=0;
    for(i=0; b[nu]<=x&&i<nu; i++)
    {
        if(x%b[i]==0)
        {
            c[cn][0]=b[i];
            while(x%b[i]==0)
            {
                c[cn][1]++;
                x/=b[i];
            }
            cn++;
        }
    }
    if(x!=1)
    {
        c[cn][0]=x;
        c[cn++][1]=1;
    }
}
long long so(long long n,long long x)
{
    long long i=0,sum=n;
    while(x)
    {
        if(x&1)
        {
            sum=sum/c[i][0]*(c[i][0]-1);
            sum*=c[i][1];
        }
        i++;
        x>>=1;
    }
    return sum;
}
long long work(long long n)
{
   long long i,size=1<<cn,sum=0;
   for(i=0;i<size;i++)
   {
       sum+=so(n,i);
   }
   return sum;
}
int main()
{
    init();
    long long n;
    while(~scanf("%I64d",&n))
    {
        fun(n);
        printf("%I64d
",work(n));
    }
}

设函数g(n) = gcd(i,n) (1<=i<=n)，由积性函数的定义，g(n)=g(m1)*g(m2) (n=m1*m2 且 （m1, m2）= 1)，所以g是积性函数。由具体数学上的结论，积性函数的和也是积性的。所以f(n) = ∑gcd(i, n)也是积性函数。由初等数论中的定理，如果f(n)是不恒为0的数论函数，n>1时 n=p1^a1*p2^a2*...*ps^as，那么f(n)是积性函数的充要条件是f(1)=1，及f(n) = f(p1^a1)*f(p2^a2)*...f(pr^ar)。所以只要求f(pi^ai)就好，如果d是n的一个约数，那么1<=i<=n中gcd(i,n) = d的个数是phi(n/d)，即n/d的欧拉函数

f(pi^ai) =  Φ（pi^ai）+pi*Φ（pi^(ai-1)）+pi^2*Φ（pi^(ai-2)）+...+pi^(ai-1)* Φ（pi）+ pi^ai *Φ（1）

     = pi^(ai-1)*(pi-1) + pi*pi^(ai-2)*(pi-1)....+pi^ai

     =  pi^ai*(1+ai*(1-1/pi))

f(n) = p1^a1*p2^a2...*pr^ar*(1+a1*(1-1/p1))*(1+a2*(1-1/p2))*...

       =  n*(1+a1*(1-1/p1))*(1+a2*(1-1/p2))*...

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <string.h>
#define MAXN 100010
#define INF 0x7fffffff
using namespace std;
int a[50000]= {0},b[50000],nu,c[70][2],cn;
void init()
{
    nu=0;
    long long i,j;
    for(i=2; i<50000; i++)
    {
        if(!a[i])
        {
            b[nu++]=i;
            j=i*i;
            while(j<50000)
            {
                a[j]=1;
                j+=i;
            }
        }
    }
}
void fun(long long x)
{
    long long i;
    memset(c,0,sizeof(c));
    cn=0;
    for(i=0; b[nu]<=x&&i<nu; i++)
    {
        if(x%b[i]==0)
        {
            c[cn][0]=b[i];
            while(x%b[i]==0)
            {
                c[cn][1]++;
                x/=b[i];
            }
            cn++;
        }
    }
    if(x!=1)
    {
        c[cn][0]=x;
        c[cn++][1]=1;
    }
}
long long work(long long n)
{
   long long i,sum=n;
   for(i=0;i<cn;i++)
   {
       sum=sum+sum*c[i][1]*(c[i][0]-1)/c[i][0];
   }
   return sum;
}
int main()
{
    init();
    long long n;
    while(~scanf("%I64d",&n))
    {
        fun(n);
        printf("%I64d
",work(n));
    }
}