POJ 2480 Longge's problem （欧拉函数）

Longge’s problem

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 8107 Accepted: 2680

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

思路：根据N的范围可知暴力枚举肯定是不行的了，我们可以通过N的约数来求出gcd和，通过枚举N的约数的欧拉函数值及其商的欧拉函数值相乘求出，注意当N可以开方的时候，减去重复运算的次数，复杂度为O(N−−√).

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
ll eular(ll n)
{
ll ret=1,i;
for(i=2;i*i<=n;i++)
{
if(n%i==0)
{
n/=i,ret*=i-1;
while(n%i==0)
n/=i,ret*=i;
}
}
if(n>1)
ret*=n-1;
return ret;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
ll ans=0;
for(int i=1;i<=(int)sqrt(n*1.0);i++)
{
if(n%i==0)
{
ans+=n/i*eular(i);
if(i*i!=n)
ans+=i*eular(n/i);
}
}
printf("%I64d\n",ans);
}
return 0;
}