zoj 3175 Number of Containers (大数灵活题，除2改成平方根)

For two integers m and k, k is said to be a container of m if k is divisible by m. Given 2 positive integers n and m (m < n), the function f(n, m) is
defined to be the number of containers of m which are also no greater than n. For example, f(5, 1)=4, f(8, 2)=3, f(7, 3)=1, f(5, 4)=0...

Let us define another function F(n) by the following equation:



Now given a positive integer n,
you are supposed to calculate the value of F(n).

Input

There are multiple test cases. The first line of input contains an integer T(T<=200) indicating the number of test cases. Then Ttest cases follow.

Each test case contains a positive integer n (0 < n <= 2000000000) in a single line.

Output

For each test case, output the result F(n) in a single line.

Sample Input

2
1
4

Sample Output

0
4
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3216

本质就是求(n/1+n/2+...+n/n-1)-n的值
数字太大，不能用n/2,只能用平方根（这个在比赛的时候完全想不到。。）
推导过程：比如 n=10  
	=1*10
	=2*5
	=3*3
10-5=5,5-3=2
然后我们又惊奇地发现最后的结果10,5,3,2,2,1,1,1,1,1
1有5个，2有2个，于是找到规律。
但是对于像8这样的数字来说，还有两个数字没有赋值到，于是sum+=未赋值个数*当前的b
其他方法：/article/2411221.html

int main(){
	int t;
	cin>>t;
	while(t--){
		ll n;
		cin>>n;
		ll s=0;
		ll a=n,b=1,c=1;
		for(ll i=2;i*i<=n;++i){
			s+=n/i+(a-n/i)*b;
			c+=a-(n/i)+1;
			b++;
			a=n/i;
		}
		s+=(n-c)*b;
		cout<<s<<endl;
	}
	return 0;
}