POJ-2480 Longge's problem 积性函数

  题目链接:http://poj.org/problem?id=2480

  题意:多次求sigma(gcd(i,n), 1<=i<=n<2^32)

  这题不能直接搜了,需要考虑函数的性质。

  由gcd(i*j,n)=gcd(i,n)*gcd(j,n),所以gcd(i,n)为积性函数。

  设f(n)=Σ(gcd(i,n)),由定理:积性函数的和函数也是积性函数(具体数学上有)。

  所以f(x)=f(p1^a1*p2^a1*...*pn^an)=f(p1^a1)*f(p2*a2)*...*f(pn^an)。

  只要对每个f(pi^ai)求解就可以了,f(pi^ai)=1*phi(pi^ai)+pi^a1*phi(pi^(ai-1))+...+pi^ai*phi(1)。

  由phi(pi^ai) = pi^ai - pi^(ai-1),那么可以化简上面的式子:f(pi^ai) = ai * pi^ai - ai * pi^(ai-1) + pi^ai = pi^ai * (ai - ai/pi + 1);

 1 //STATUS:C++_AC_47MS_116KB
 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 __int64 LL;
34 typedef unsigned __int64 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 ans,n;
59 
60 int main(){
61  //   freopen("in.txt","r",stdin);
62     LL i,j;
63     LL cnt,t;
64     while(~scanf("%I64d",&n)){
65         ans=n;t=1;
66         for(i=2;i*i<=n;i++){
67             if(n%i==0){
68                 cnt=0 ;
69                 while(n%i==0){
70                     cnt++;
71                     n/=i;
72                 }
73                 ans+=ans*cnt/i*(i-1);
74             }
75         }
76         if(n>1)ans=ans*(n*2-1)/n;
77 
78         printf("%I64d
",ans);
79     }
80     return 0;
81 }
原文地址:https://www.cnblogs.com/zhsl/p/3281785.html