Bzoj-2705 Longge的问题 欧拉函数

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

　　题意： 求 sigma(gcd(i,n), 1<=i<=n<2^32)

　　只有一组数据，很好搞，答案就是sigma(phi(n/d))，直接搜就行了。

 //STATUS:C++_AC_8MS_11284KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 //#include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef long long LL;
 typedef unsigned long long ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End LL p[N][];
 LL ans,n,cnt; void dfs(LL d,LL phi)
 {
     if(d==cnt){
         ans+=phi;
         return ;
     }
     dfs(d+,phi);
     phi=phi/p[d][]*(p[d][]-);
     for(int i=;i<=p[d][];i++)
         dfs(d+,phi);
 } int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,la;
     LL t;
     scanf("%lld",&n);
     cnt=;
     for(t=n,i=;i*i<=t;i++){
         if(t%i==){
             p[cnt][]=i;
             while(t%i==){
                 p[cnt][]++;
                 t/=i;
             }
             cnt++;
         }
     }
     if(t)p[cnt][]=t,p[cnt][]=,cnt++;     ans=;
     dfs(,n);     printf("%lld\n",ans);
     return ;
 }