更准确的代码！！！

2022-11-27 09:14:22

发布于：江苏

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首先，暴力枚举直到n肯定不可以，因为会TLE
那我们要限定枚举范围
取到哪里呢？
首先我们要知道，除了完全平方数以外，其他的数中，约数是成对出现的，因此直接枚举到其平方根
然后判断如果是输入的数的平方根，只+1
上！代码！

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n,ans = 0;
	cin>>n;
	for (int i=1;i<=sqrt(n);i++)
		if (n%i== 0){
			if (n/i==i){
				ans++;
				continue;
			}
		ans += 2;
	}
	if (n==1) ans=1;
	cout<<ans;
	return 0; //&lt;要替换成一个小于号，然后进行阅读
}

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f u c k ccf
为什么要continue

2024-06-26 来自广东
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2026-02-11 来自浙江
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2024-08-17 来自江苏
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