}LLlehmer_pi(LL x){//小于等于n的素数有多少个if(x < N)returnpi[x];inta = (int)lehmer_pi(sqrt2(sqrt2(x)));intb = (int)lehmer_pi(sqrt2(x));intc = (int)lehmer_pi(sqrt3(x)); LL sum =getphi(x, a) +(LL)(b + a -2) * (b - a +1) /2;for(inti = a +1; i ...
}LLlehmer_pi(LL x){//小于等于n的素数有多少个if(x < N)returnpi[x];inta = (int)lehmer_pi(sqrt2(sqrt2(x)));intb = (int)lehmer_pi(sqrt2(x));intc = (int)lehmer_pi(sqrt3(x)); LL sum =getphi(x, a) +(LL)(b + a -2) * (b - a +1) /2;for(inti = a +1; i ...
LLlehmer_pi(LLx) { if(x<N)returnpi[x]; inta=(int)lehmer_pi(sqrt2(sqrt2(x))); intb=(int)lehmer_pi(sqrt2(x)); intc=(int)lehmer_pi(sqrt3(x)); LLsum=getphi(x,a)+LL(b+a-2)*(b-a+1)/2; for(inti=a+1;i<=b;i++) { LLw=x/prime[i]; sum-=lehmer_pi(w); if(i...
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 = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2; for (int i = a + 1; ...
Meisell-Lehmer算法(统计较大数据里的素数) http://acm.hdu.edu.cn/showproblem.php?pid=5901 1e11的数据量,这道题用这个算法花了202ms. 1#include<bits/stdc++.h>23usingnamespacestd;45typedeflonglongLL;6constintN = 5e6 +2;7boolnp[N];8intprime[N], pi[N];910intgetprime() {11intcnt =...
[i]) - i +1;69returnans;70}7172LL lehmer_pi(LL x){73if(x < N)returnpi[x];74inta = (int)lehmer_pi(sqrt2(sqrt2(x)));75intb = (int)lehmer_pi(sqrt2(x));76intc = (int)lehmer_pi(sqrt3(x));77LL sum = getphi(x, a) +(LL)(b + a -2) * (b - a +1) /2;78...
HDU 5901 Count primes( Meisell-Lehmer算法模板 ) 链接:***传送门 题意:计算 [ 1 , n ] 之间素数的个数,(1 <= n <= 1e11) 思路:Meisell-Lehmer算法是计算超大范围内素数个数的一种算法,原理并不明白,由于英语太渣看不懂WIKI上的原理,附WIKI链接:Here ...