But Chau Kok screen hit the table or overtime, hit the table long enough it will have to run in 51nod top 5s. If you want to get close to 10 times faster 1000 times, do not come out to play (meaning time).
Prime violent check looking for is not to be one.
#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int MAXN = 7500000;
int n;
int f[MAXN + 5], p[MAXN + 5];
bool np[MAXN + 5];
ll g(ll n, int m) {
if(!m)
return n;
if(m == 1)
return n - n / 2;
if(n <= MAXN) {
if(f[n] <= m)
return 1;
if(f[(int)sqrt(n)] <= m)
return f[n] - m + 1;
}
ll res = g(n, m - 1) - g(n / p[m], m - 1);
return res;
}
bool pan(ll x) {
ll y = sqrt(x);
return f[y] + g(x, f[y]) - 1 >= n;
}
void init() {
for(int i = 2; i <= MAXN; ++i) {
if(!np[i])
p[++p[0]] = i;
for(int j = 1, t; j <= p[0] && (t = p[j] * i) <= MAXN; ++j) {
np[t] = 1;
if(i % p[j] == 0)
break;
}
}
for(int i = 2; i <= MAXN; ++i)
f[i] = f[i - 1] + (np[i] == 0);
}
ll pcount[] = {
373587883,
776531401,
1190494759,
1611623773,
2038074743,
2468776129,
2902958801,
3340200037,
3780008329,
4222234741,
4666527007,
5112733757,
5560695863,
6010236857,
6461335109,
6913774603,
7367575799,
7822624247,
8278737359,
8736028057,
9194418049,
9653704481,
10113958157,
10575209467,
11037271757,
11500205947,
11963902331,
12428375423,
12893587657,
13359555403,
13826206699,
14293566641,
14761538761,
15230122499,
15699342107,
16169207209,
16639648327,
17110593779,
17582163853,
18054236957,
18526876243,
18999999247,
19473535801,
19947663787,
20422213579,
20897216723,
21372698029,
21848603809,
22325014259,
22801763489
};
int main() {
#ifdef Yinku
freopen("Yinku.in", "r", stdin);
//freopen("Yinku.out", "w", stdout);
#endif // Yinku
init();
while(~scanf("%d", &n)) {
int px = n / (int)(2e7);
ll l = px == 0 ? 1 : pcount[px - 1], r = pcount[px], mid;
//cout << "l=" << l << " r=" << r << endl;
while(l < r) {
mid = (l + r) / 2;
if(pan(mid))
r = mid;
else
l = mid + 1;
}
printf("%lld\n", l);
}
}The following good confuse ah;
#include<iostream>
#include<cmath>
#include<assert.h>
using namespace std;
typedef long long LINT;
LINT a,b,goal,n;
int mark[160000],prime[160000],e,bl[10000005];
LINT pn(int n)
{
LINT s=LINT(n*(log(n)+log(log(n))-1)+n*(log(log(n))-2)/log(n)-6.0*n/1000.0);
return s<=1?1:s;
}
inline LINT V2IDX(LINT v, LINT N, LINT Ndr, LINT nv) {
return v >= Ndr ? (N/v - 1) : (nv - v);
}
LINT primesum(LINT N) {
LINT *S;
LINT *V;
LINT r = (LINT)sqrt(N);
LINT Ndr = N/r;
assert(r*r <= N and (r+1)*(r+1) > N);
LINT nv = r + Ndr - 1;
V = new LINT[nv];
S = new LINT[nv];
for (LINT i=0; i<r; i++) {
V[i] = N/(i+1);
}
for (LINT i=r; i<nv; i++) {
V[i] = V[i-1] - 1;
}
for (LINT i=0; i<nv; i++) {
//S[i] = V[i] * (V[i] + 1) / 2 - 1;若求素数和，使用此处
S[i]=V[i] - 1;
//若求素数个数使用此处
}
for (LINT p=2; p<=r; p++) {
if (S[nv-p] > S[nv-p+1]) {
LINT sp = S[nv-p+1];
LINT p2 = p*p;
for (LINT i=0; i<nv; i++) {
if (V[i] >= p2) {
//S[i] -= p* (S[V2IDX(V[i]/p, N, Ndr, nv)] - sp);若求素数和，使用此处
S[i] -= 1* (S[V2IDX(V[i]/p, N, Ndr, nv)] - sp);
//若求素数个数，使用此处
} else {
break;
}
}
}
}
return S[0];
}
int main()
{
cin>>n;
a=pn(n);
if(a%2&&a>1)a=a-1;//防止预估值本身就是素数的情况，刚开始被这里坑了3个test
b=a+7000000;
goal=n-primesum(a);
for(int i=2;i<=160000;i++)
{
if(!mark[i])
{
prime[++e]=i;
for(LINT j=(LINT)i*i;j<=160000;j+=i)
{
mark[j]=1;
}
}
}
LINT xxx,c;
for(int i=1;i<=e;i++)
{
xxx=(LINT)ceil(1.0*a/prime[i]);
if(xxx==1) xxx++;
for(LINT j=xxx;(c=j*prime[i])<b;j++)
{
bl[c-a]=1;
}
}
int ans=0;
c=b-a;
if(a==1) ans--;
for(int i=0;i<c;i++)
{
if(!bl[i]) ans++;
if(ans==goal)
{
cout<<i+a<<endl;
break;
}
}
return 0;
}Meisell-Lehmer algorithm calculates the number of prime numbers between 2 ~ n and half on it.
bool np[maxn];
int prime[maxn],pi[maxn];
int getprime()
{
int cnt=0;
np[0]=np[1]=true;
pi[0]=pi[1]=0;
for(int i=2; i<maxn; ++i)
{
if(!np[i]) prime[++cnt]=i;
pi[i]=cnt;
for(int j=1; j<=cnt&&i*prime[j]<maxn; ++j)
{
np[i*prime[j]]=true;
if(i%prime[j]==0) break;
}
}
return cnt;
}
const int M=7;
const int PM=2*3*5*7*11*13*17;
int phi[PM+1][M+1],sz[M+1];
void init()
{
getprime();
sz[0]=1;
for(int i=0; i<=PM; ++i) phi[i][0]=i;
for(int i=1; i<=M; ++i)
{
sz[i]=prime[i]*sz[i-1];
for(int j=1; j<=PM; ++j)
{
phi[j][i]=phi[j][i-1]-phi[j/prime[i]][i-1];
}
}
}
int sqrt2(ll x)
{
ll r=(ll)sqrt(x-0.1);
while(r*r<=x) ++r;
return int(r-1);
}
int sqrt3(ll x)
{
ll r=(ll)cbrt(x-0.1);
while(r*r*r<=x) ++r;
return int(r-1);
}
ll getphi(ll x,int s)
{
if(s==0) return x;
if(s<=M) return phi[x%sz[s]][s]+(x/sz[s])*phi[sz[s]][s];
if(x<=prime[s]*prime[s]) return pi[x]-s+1;
if(x<=prime[s]*prime[s]*prime[s]&&x<maxn)
{
int s2x=pi[sqrt2(x)];
ll ans=pi[x]-(s2x+s-2)*(s2x-s+1)/2;
for(int i=s+1; i<=s2x; ++i)
{
ans+=pi[x/prime[i]];
}
return ans;
}
return getphi(x,s-1)-getphi(x/prime[s],s-1);
}
ll getpi(ll x)
{
if(x<maxn) return pi[x];
ll ans=getphi(x,pi[sqrt3(x)])+pi[sqrt3(x)]-1;
for(int i=pi[sqrt3(x)]+1,ed=pi[sqrt2(x)]; i<=ed; ++i)
{
ans-=getpi(x/prime[i])-i+1;
}
return ans;
}
ll lehmer_pi(ll x)
{
if(x<maxn) return pi[x];
int a=(int)lehmer_pi(sqrt2(sqrt2(x)));
int b=(int)lehmer_pi(sqrt2(x));
int c=(int)lehmer_pi(sqrt3(x));
ll sum=getphi(x,a)+ll(b+a-2)*(b-a+1)/2;
for(int i=a+1; i<=b; i++)
{
ll w=x/prime[i];
sum-=lehmer_pi(w);
if(i>c) continue;
ll lim=lehmer_pi(sqrt2(w));
for(int j=i; j<=lim; j++)
{
sum-=lehmer_pi(w/prime[j])-(j-1);
}
}
return sum;
}