注意这道题开得非常大,有2*1e7 自己可以养成一种习惯,如果数据是很容易的话,可以自己手动输入极限数据来测试自己的程序 #include<cstdio> #include<algorithm> #include<cstring> #include<vector> #define REP(i, a, b) for(int i = (a); i < (b); i++) #define _for(i, a, b) for(int i = (a); i &l…
O(n)线性筛选n以内的素数 (1)对于任何一个素数p,都不可能表示为两个数的乘积 (2)对于任何一个合数m = p1a1p2a2…pmam,这里p1< p2 < … <pm,都能使用p1a1-1p2a2…pmam* p1进行筛选 fillchar(prime,sizeof(prime),); prime[]:=false; fillchar(p,sizeof(p),); total:=; to n do begin if prime[i] then begin inc(total);p…
Problem Description Goldbach's Conjecture: For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2. This conjecture has not been proved nor refused yet. No one is sure whether…
Description: Count the number of prime numbers less than a non-negative number, n. 题解:就是线性筛素数的模板题. class Solution { public: int countPrimes(int n) { ; vector<,); ;i<n;i++){ if(is_prime[i]){ ans++; *i;j<n;j+=i){ is_prime[j]=; } } } return ans; } }…