[推销博客](https://foreverlasting1202.github.io/2019/08/27/CF1208%E9%A2%98%E8%A7%A3/)
### G Polygons
题意：给出$n$和$k$，要求在圆上用最少的点数使得能够找到$k$个正多边形而且这$k$个正多边形的长度小于等于$n$，求最少的点数。$1\leq k\leq n-2\leq 10^6$。
做法：考虑如果选择一个数$x$（不考虑$1$和$2$的情况），则显然与它不互质的数都已经选过了（否则选不互质的数会更优），那么这个数的贡献就会是$\phi(x)$。因此给$\phi$排个序，找前$k$个就可以了。时间复杂度$O(nlogn)$。
code:
```cpp
//2019.8.27 by ljz
//email
[email protected]
//if you find any bug in my code
//please tell me
#include<bits/stdc++.h>
using namespace std;
#define res register int
#define LL long long
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f
#define unl __int128
#define eps 5.6e-8
#define RG register
#define db double
#define pc(x) __builtin_popcount(x)
//#define pc(x) __builtin_popcountll(x)
typedef pair<int,int> Pair;
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pb push_back
#define ull unsigned LL
#define gc getchar
//inline char gc() {
// static char buf[100000],*p1,*p2;
// return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
//}
inline int read() {
res s=0,ch=gc();
while(ch<'0'||ch>'9')ch=gc();
while(ch>='0'&&ch<='9')s=s*10+ch-'0',ch=gc();
return s;
}
//inline int read() {
// res s=0,ch=gc(),w=1;
// while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=gc();}
// while(ch>='0'&&ch<='9')s=s*10+ch-'0',ch=gc();
// return s*w;
//}
//inline LL Read() {
// RG LL s=0;
// res ch=gc();
// while(ch<'0'||ch>'9')ch=gc();
// while(ch>='0'&&ch<='9')s=s*10+ch-'0',ch=gc();
// return s;
//}
//inline LL Read() {
// RG LL s=0;
// res ch=gc(),w=1;
// while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=gc();}
// while(ch>='0'&&ch<='9')s=s*10+ch-'0',ch=gc();
// return s*w;
//}
//inline void write(RG unl x){
// if(x>10)write(x/10);
// putchar(int(x%10)+'0');
//}
inline void swap(res &x,res &y) {
x^=y^=x^=y;
}
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//clock_t start=clock();
//inline void ck(){
// if(1.0*(clock()-start)/CLOCKS_PER_SEC>0.1)exit(0);
//}
const int N=1e6+10;
namespace MAIN{
int n,k;
int prim[N/10],tot,phi[N];
bool vis[N];
inline void MAIN(){
phi[1]=1,n=read(),k=read();
for(res i=2;i<=n;i++){
if(!vis[i])prim[++tot]=i,phi[i]=i-1;
for(res j=1;j<=tot&&prim[j]*i<=n;j++){
vis[prim[j]*i]=1;
if(i%prim[j]==0){phi[prim[j]*i]=phi[i]*prim[j];;break;}
phi[prim[j]*i]=phi[i]*(prim[j]-1);
}
}
sort(phi+1,phi+n+1);
RG LL ans=0;
for(res i=1;i<=k+2;i++)ans+=phi[i];
printf("%lld\n",ans-(k==1));
}
}
int main(){
// srand(19260817);
// freopen("signin.in","r",stdin);
// freopen("signin.out","w",stdout);
MAIN::MAIN();
return 0;
}
```