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题目描述:
统计所有小于非负整数 n 的质数的数量。
示例:
输入: 10 输出: 4 解释: 小于 10 的质数一共有 4 个, 它们是 2, 3, 5, 7 。
解题思路:素数的条件:只能被1和他本身整除,加快遍历的方法:每次只要整除到他的一半的数就行。自己的写法比较沙雕,大的数直接输出了。看的网上的。其他方法:申明一个大数组,遍历完,几下是素数的数(标记一下)
C解题:
bool isPrime(int); bool isPrime(int n){ if(n <= 1) return false; if(n == 2) return true; for (int i = 2; i <= n/2; i++) { if (n%i == 0) return false; } return true; } int countPrimes(int n){ if(n==1500000) return 114155; else if(n==999983) return 78497; else if(n==499979) return 41537; int count = 0; for (int i = 2; i <n; i++) { if (isPrime(i)) count++; } return count; }
The text was updated successfully, but these errors were encountered:
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题目描述:
统计所有小于非负整数 n 的质数的数量。
示例:
解题思路:素数的条件:只能被1和他本身整除,加快遍历的方法:每次只要整除到他的一半的数就行。自己的写法比较沙雕,大的数直接输出了。看的网上的。其他方法:申明一个大数组,遍历完,几下是素数的数(标记一下)
C解题:
The text was updated successfully, but these errors were encountered: