Description
有一个长度为N的环，上面写着’X’和’E’，问本质不同的环有多少种。(N不超过200000)。
Input
The input file contains a single integer 1 <= n <= 200000.
Output
Output a single integer --- the number circular strings of length n.
Sample Input
3
Sample Output
4
polya裸题，和pku2154 Color一样，只是需要带个高精度罢了，只有两种颜色，不用取模
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
int x=0,f=1;char ch=getchar();
for (;ch<‘0‘||ch>‘9‘;ch=getchar()) if (ch==‘-‘) f=-1;
for (;ch>=‘0‘&&ch<=‘9‘;ch=getchar()) x=(x<<1)+(x<<3)+ch-‘0‘;
return x*f;
}
inline void print(int x){
if (x>=10) print(x/10);
putchar(x%10+‘0‘);
}
const int N=2e5;
const int base=1e4;
const int digit=4;
const int maxn=1e2;
struct Bignum{
int len,v[maxn];
Bignum(){len=1,memset(v,0,sizeof(v));}
void init(){v[0]=1;}
void write(){
printf("%d",v[len-1]);
for (int i=len-2;~i;i--) printf("%0*d",digit,v[i]);
putchar(‘\n‘);
}
};
int prime[N+10],phi[N+10];
bool inprime[N+10];
int tot;
void prepare(){
phi[1]=1;
for (int i=2;i<=N;i++){
if (!inprime[i]) prime[++tot]=i,phi[i]=i-1;
for (int j=1;j<=tot&&prime[j]*i<=N;j++){
inprime[prime[j]*i]=1;
if (i%prime[j]==0) phi[i*prime[j]]=phi[i]*prime[j];
else phi[i*prime[j]]=phi[i]*(prime[j]-1);
}
}
}
Bignum operator +(const Bignum &x,const Bignum &y){
Bignum z;
z.len=max(x.len,y.len);
for (int i=0;i<=z.len;i++) z.v[i]+=x.v[i]+y.v[i],z.v[i+1]+=z.v[i]/base,z.v[i]%=base;
while (z.v[z.len]) z.v[z.len+1]+=z.v[z.len]/base,z.v[z.len]%=base,z.len++;
return z;
}
Bignum operator *(const Bignum &x,const Bignum &y){
Bignum z;
z.len=x.len+y.len;
for (int i=0;i<=x.len;i++)
for (int j=0;j<=y.len;j++)
z.v[i+j]+=x.v[i]*y.v[j],z.v[i+j+1]+=z.v[i+j]/base,z.v[i+j]%=base;
while (z.len!=1&&!z.v[z.len]) z.len--;z.len++;
while (z.v[z.len]) z.v[z.len+1]+=z.v[z.len]/base,z.v[z.len]%=base,z.len++;
return z;
}
Bignum operator /(Bignum &x,int y){
for (int i=x.len;~i;i--) x.v[i-1]+=x.v[i]%y*base,x.v[i]/=y;
while (x.len!=1&&!x.v[x.len]) x.len--;
while (x.v[x.len]) x.v[x.len+1]+=x.v[x.len]/base,x.v[x.len]%=base,x.len++;
return x;
}
Bignum mlt(Bignum a,int b){
Bignum res;res.init();
for (;b;b>>=1,a=a*a) if (b&1) res=res*a;
return res;
}
Bignum change(int x){
Bignum z;
while (x){
z.v[z.len-1]=x%base;
z.len++;
x/=base;
}
while (z.len!=1&&!z.v[z.len]) z.len--;
while (z.v[z.len]) z.v[z.len+1]+=z.v[z.len]/base,z.v[z.len]%=base,z.len++;
return z;
}
int main(){
prepare();
int n=read();
Bignum Two,Ans;
Two.v[0]=2;
for (int i=1;i*i<=n;i++){
if (n%i) continue;
int j=n/i;
Ans=Ans+mlt(Two,i)*change(phi[j]);
if (i!=j) Ans=Ans+mlt(Two,j)*change(phi[i]);
}
Ans=Ans/n;
Ans.write();
return 0;
}