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。。

 1 //STATUS:C++_AC_24MS_1584KB

 2 #include <functional>

 3 #include <algorithm>

 4 #include <iostream>

 5 //#include <ext/rope>

 6 #include <fstream>

 7 #include <sstream>

 8 #include <iomanip>

 9 #include <numeric>

10 #include <cstring>

11 #include <cassert>

12 #include <cstdio>

13 #include <string>

14 #include <vector>

15 #include <bitset>

16 #include <queue>

17 #include <stack>

18 #include <cmath>

19 #include <ctime>

20 #include <list>

21 #include <set>

22 #include <map>

23 using namespace std;

24 //#pragma comment(linker,"/STACK:102400000,102400000")

25 //using namespace __gnu_cxx;

26 //define

27 #define pii pair<int,int>

28 #define mem(a,b) memset(a,b,sizeof(a))

29 #define lson l,mid,rt<<1

30 #define rson mid+1,r,rt<<1|1

31 #define PI acos(-1.0)

32 //typedef

33 typedef long long LL;

34 typedef unsigned long long ULL;

35 //const

36 const int N=40010;

37 const int INF=0x3f3f3f3f;

38 const int MOD=100000,STA=8000010;

39 const LL LNF=1LL<<60;

40 const double EPS=1e-8;

41 const double OO=1e15;

42 const int dx[4]={-1,0,1,0};

43 const int dy[4]={0,1,0,-1};

44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

45 //Daily Use ...

46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

56 //End

57  

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

59 int cnt;

60  

61 void phitable(int n)

62 {

63     int i,j;

64     cnt=0;phi[1]=1;

65     for(i=2;i<=n;i++){

66         if(!phi[i]){

67             prime[cnt++]=i;

68             phi[i]=i-1;

69         }

70         for(j=0;j<cnt && i*prime[j]<=n;j++){

71             if(i%prime[j]){

72                 phi[i*prime[j]]=phi[i]*(prime[j]-1);

73             }else {phi[i*prime[j]]=phi[i]*prime[j];break;}

74         }

75     }

76 }

77  

78 int n;

79  

80 int main(){

81  //   freopen("in.txt","r",stdin);

82     int i,j,ans=0;

83     scanf("%d",&n);

84     phitable(n);

85     for(i=1;i<n;i++)ans+=phi[i];

86     printf("%d\n",ans<<1|1);

87  

88     return 0;

89 }