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。

 //STATUS:C++_AC_8MS_11284KB
#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 LL p[N][];
LL ans,n,cnt; void dfs(LL d,LL phi)
{
if(d==cnt){
ans+=phi;
return ;
}
dfs(d+,phi);
phi=phi/p[d][]*(p[d][]-);
for(int i=;i<=p[d][];i++)
dfs(d+,phi);
} int main(){
// freopen("in.txt","r",stdin);
int i,j,la;
LL t;
scanf("%lld",&n);
cnt=;
for(t=n,i=;i*i<=t;i++){
if(t%i==){
p[cnt][]=i;
while(t%i==){
p[cnt][]++;
t/=i;
}
cnt++;
}
}
if(t)p[cnt][]=t,p[cnt][]=,cnt++; ans=;
dfs(,n); printf("%lld\n",ans);
return ;
}
上一篇:java socket 多线程通讯 使用mina作为服务端


下一篇:Libcurl的编译_HTTP/HTTPSclient源代码演示样例