ZOJ

B - Number of Containers

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.

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Output

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

Sample Input

2
1
4

Sample Output

0
4
运行一下 输入俩1000 即可发现新大陆

#include<bits/stdc++.h>
using namespace std;
int ans[3000000];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        long long answer=0;
        scanf("%d",&n);
        if(n==2||n==3)
        {
            cout<<n-1<<endl;
            continue;
        }
        int sq=sqrt(n);
        for(int i=2;i<=sq;i++)//反向求和
        {
            ans[i]=n/i;
            if(i==2)
            {
                continue;
            }
            cout<<'x'<<"的范围为"<<'['<<ans[i]+1<<','<<ans[i-1]<<']'<<"使"<<"f("<<n<<','<<'x'<<")="<<i-2<<"成立"<<endl;
            answer+=(ans[i-1]-ans[i])*(i-2);
        }
        for(int i=1;i<=ans[sq];i++)//反向求和
        {
            //正向求f（n，x）
            cout<<"f("<<n<<','<<i<<")="<<' '<<(n/i)-1<<endl;
            answer+=(n/i)-1;
        }
        cout<<answer<<endl;
    }
    return 0;
}