POJ

题目：给出N，求 ∑gcd(i, N) 1<=i <=N.

0<N<2^31

思路：

phi(p^k)=(p-1)*p^(k-1)

考虑如果N=P^k的时候，那么F[N]=k*p^(k-1)*(p-1)+p^k。

利用eular的积性性质

F[N]=F[p1^k1*p2^k2……pi^ki]=∑(ki*pi^(ki-1)+p^ki).

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }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;}// 0x3f3f3f3f3f3f3f3fint main(){    LL n;    while(~scanf("%lld",&n)){        LL ans=1;        for(LL i=2;i*i<=n;i++){            LL p=1,k=0;            while(n%i==0){                p*=i;                k++;                n/=i;            }            ans*=k*(p-p/i)+p;        }        if(n>1)            ans*=2*n-1;        printf("%lld\n",ans);    }    return 0;}