Bzoj-2190 仪仗队 欧拉函数

　　题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2190

　　简单的欧拉函数题，实际上就是求gcd(x,y)=1, 0<=x,y<=n的对数。。

 //STATUS:C++_AC_24MS_1584KB
 #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

 int phi[N],prime[N];
 int cnt;

 void phitable(int n)
 {
     int i,j;
     cnt=;phi[]=;
     for(i=;i<=n;i++){
         if(!phi[i]){
             prime[cnt++]=i;
             phi[i]=i-;
         }
         for(j=;j<cnt && i*prime[j]<=n;j++){
             if(i%prime[j]){
                 phi[i*prime[j]]=phi[i]*(prime[j]-);
             }else {phi[i*prime[j]]=phi[i]*prime[j];break;}
         }
     }
 }

 int n;

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,ans=;
     scanf("%d",&n);
     phitable(n);
     for(i=;i<n;i++)ans+=phi[i];
     printf("%d\n",ans<<|);

     return ;
 }