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;}