http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3175

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 T test 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

 

题意：求n/1+n/2+n/3+……+n/n-n的值

思路：n很大，只有优化到根号N才能做，怎么优化呢？

题解：https://www.cnblogs.com/ZP-Better/p/4639596.html

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
#include<map>
#include<vector>
#include<set>
#include<list>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
    	ll n;
    	scanf("%lld",&n);
    	ll q=(ll)sqrt(n);
    	ll sum=0;
    	for(int i=1;i<=q;i++){
    		sum+=n/i;
		}
		printf("%lld\n",sum*2-q*q-n);
	}
	return 0;
}

分块

#include<cstdio>
using namespace std;
typedef long long ll;
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int n;
		scanf("%d",&n);
		ll ans=0;
		for(int l=1,r;l<=n;l=r+1){
			r=n/(n/l);
			ans+=(ll)(n/l)*(r-l+1);
		}
		printf("%lld\n",ans-n);
	}
	return 0;
}