1 class Solution 2 { 3 public: 4 int countPrimes(int n) 5 { 6 vector<bool> isPrime(n+1,true); //首先全部为true 7 int count = 0; 8 9 for(int i = 2; i < n;++i)//从2开始 10 { 11 if(isPrime[i]) 12 { 13 ++count; 14 for(int j = 2 ; i * j < n ; ++j)//能被整除的全部筛掉,置为false 15 { 16 isPrime[i * j] = false; 17 } 18 } 19 } 20 return count; 21 } 22 23 };
原文:https://www.cnblogs.com/yuhong1103/p/12632148.html