ZOJ 3175 Number of Containers

ZOJ Problem Set - 3175

Number of Containers

Time Limit: 1 Second      Memory Limit: 32768 KB

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

Author: CAO, Peng

Contest: The 9th Zhejiang University Programming Contest

n*(1/1+1/2+1/3+1/4+.........+1/n)-n
枚举到sqrt（n）×2-重复部分

#include <iostream>

#include <cstring>

#include <cstdio>

#include <cmath>

using namespace std;

typedef unsigned long long int uLL;

int main()

{

    int T;

    int n;

    uLL ans;

    cin>>T;

while(T--)

{

    cin>>n;

    int nt=(int)sqrt(n*1.0);

    ans=0;

    for(int i=1;i<=nt;i++)

    {

        ans+=n/i;

    }

    ans=ans*2-nt*nt;

    ans-=n;

    cout<<ans<<endl;

}

    return 0;

}