# Count primes

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 635    Accepted Submission(s): 299

Problem Description
Easy question! Calculate how many primes between [1…n]!

Input
Each line contain one integer n(1 <= n <= 1e11).Process to end of file.

Output
For each case, output the number of primes in interval [1…n]

Sample Input
2
3
10

Sample Output
1
2
4

1n(n1011) 之间有多少个素数，

My Code

/** 2016 - 09 - 19 晚上 Author: ITAK Motto: 本日的我要超出昨日的我，明日的我要胜过本日的我， 以创作出更好的代码为目标，不断地超出自己。 **/ #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <queue> #include <algorithm> #include <set> #include <time.h> using namespace std; typedef long long LL; typedef unsigned long long ULL; const LL INF = 1e9+5; const LL MAXN = 1e6+5; const LL MOD = 1e9+7; const double eps = 1e⑺; const double PI = acos(-1); using namespace std; LL Scan_LL()///输入外挂 { LL res=0,ch,flag=0; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+ch-'0'; return flag?-res:res; } void Out(LL a)///输出外挂 { if(a>9) Out(a/10); putchar(a%10+'0'); } 

const int N = 5e6 + 2; bool np[N]; int prime[N], pi[N]; int getprime() { int cnt = 0; np[0] = np[1] = true; pi[0] = pi[1] = 0; for(int i = 2; i < N; i++) { if(!np[i]) prime[++cnt] = i; pi[i] = cnt; for(int j = 1; j <= cnt && i * prime[j] < N; 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 get_phi(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 < N) { 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 get_phi(x, s - 1) - get_phi(x / prime[s], s - 1); } LL getpi(LL x) { if(x < N) return pi[x]; LL ans = get_phi(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 < N) 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 = get_phi(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; } int main() { Init(); LL n; while(~scanf("%I64d",&n)) { Out(lehmer_pi(n)); puts(""); } return 0; } 

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