Bzoj-2190 仪仗队 欧拉函数

时间：2021-12-06 16:49:05

　　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 ;

 }

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