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。

 1 //STATUS:C++_AC_8MS_11284KB
 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=110;
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 LL p[N][2];
59 LL ans,n,cnt;
60  
61 void dfs(LL d,LL phi)
62 {
63     if(d==cnt){
64         ans+=phi;
65         return ;
66     }
67     dfs(d+1,phi);
68     phi=phi/p[d][0]*(p[d][0]-1);
69     for(int i=1;i<=p[d][1];i++)
70         dfs(d+1,phi);
71 }
72  
73 int main(){
74  //   freopen("in.txt","r",stdin);
75     int i,j,la;
76     LL t;
77     scanf("%lld",&n);
78     cnt=0;
79     for(t=n,i=2;i*i<=t;i++){
80         if(t%i==0){
81             p[cnt][0]=i;
82             while(t%i==0){
83                 p[cnt][1]++;
84                 t/=i;
85             }
86             cnt++;
87         }
88     }
89     if(t)p[cnt][0]=t,p[cnt][1]=1,cnt++;
90  
91     ans=0;
92     dfs(0,n);
93  
94     printf("%lld\n",ans);
95     return 0;
96 }

 

posted @ 2013-08-26 01:31  zhsl  阅读(201)  评论(0编辑  收藏  举报