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 }
