新视野OJ 2705 [SDOI2012]Longge的问题 （数论）
传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2705
题解：求 sigma(gcd(i,n), 1<=i<=n<2^32)
又是令gcd(i, n) = d，答案就是sigma(phi(n/d))，但是我们不能预处理出phi[]数组，因为开不了数组……
注意到因数个数是O(2sqrt(n))级别的，我们枚举所有的n/d，一边dfs一边算phi。
AC代码：
2705Accepted1272 kb4 msC++/Edit
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;#define si1(a) scanf("%d",&a)#define si2(a,b) scanf("%d%d",&a,&b)#define sd1(a) scanf("%lf",&a)#define sd2(a,b) scanf("%lf%lf",&a,&b)#define ss1(s) scanf("%s",s)#define pi1(a) printf("%d\n",a)#define pi2(a,b) printf("%d %d\n",a,b)#define mset(a,b) memset(a,b,sizeof(a))#define forb(i,a,b) for(int i=a;i<b;i++)#define ford(i,a,b) for(int i=a;i<=b;i++)typedef long long LL;const int N=40000;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const double eps=1e-7;int n,cnt,p[30],c[30];LL ans=0;void dfs(int step,int pdt,int phi){ if(step==cnt) { ans+=phi; return; } dfs(step+1,pdt,phi); phi=phi/p[step]*(p[step]-1); for(int i=1;i<=c[step];++i) dfs(step+1,pdt*=p[step],phi);}int main(){ scanf("%d",&n); int x=n; for(int i=2;i*i<=x;++i) if(x%i==0) { for(;x%i==0;x/=i) ++c[cnt]; p[cnt++]=i; } if(x>1) c[cnt]=1,p[cnt++]=x; dfs(0,1,n); printf("%lld\n",ans); return 0;}