如题。$n leq 1e9$。
方法零：二分，然后洲阁筛。要魔改一下的洲阁筛。跑得慢。卡卡能过。没意思。
1 //#include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 //#include<map> 6 #include<math.h> 7 //#include<time.h> 8 //#include<complex> 9 #include<algorithm> 10 using namespace std; 11 12 #define LL long long 13 LL n,mzz=10000000; 14 #define maxn 640011 15 #define maxm 10000011 16 int prime[maxm/10],sp[maxm],lp; bool notprime[maxm]; 17 void pre() 18 { 19 lp=0; 20 for (int i=2;i<=mzz;i++) 21 { 22 if (!notprime[i]) {prime[++lp]=i;} 23 sp[i]=sp[i-1]+!notprime[i]; 24 for (int j=1,tmp;j<=lp && 1ll*prime[j]*i<=mzz;j++) 25 { 26 notprime[tmp=i*prime[j]]=1; 27 if (i%prime[j]); else break; 28 } 29 } 30 } 31 32 LL G(int i,LL j) 33 { 34 if (i==1) return (j+1)>>1; 35 if (j<=prime[i]) return 1; 36 if (j<=mzz && j<=(LL)prime[i]*prime[i]) return sp[j]+1-i; 37 return (G(i-1,j)-G(i-1,j/prime[i])); 38 } 39 40 LL mg(LL n) {return G(lp,n);} 41 42 LL calc(LL n) 43 { 44 if (n<=mzz) return sp[n]; 45 lp=1; for (;(LL)prime[lp]*prime[lp]<=n;lp++); 46 return mg(n)-1+sp[prime[lp]]; 47 } 48 49 int q; 50 int main() 51 { 52 scanf("%d",&q); pre(); 53 LL L=1,R=22801763489; 54 if (q>=900000001) L=20422213579; else R=20422213579; 55 if (q>=925000001) L=21016060633; else R=21016060633; 56 if (q>=950000001) L=21610588367; else R=21610588367; 57 if (q>=970000001) L=22086742277; else R=22086742277; 58 if (q>=985000001) L=22444149523; else R=22444149523; 59 while (L<R) 60 { 61 LL mid=(L+R)>>1,tmp; 62 if ((tmp=calc(mid))>=q) R=mid; 63 else L=mid+1; 64 } 65 printf("%lld ",L); 66 return 0; 67 }