Zoj 3175 Number of Containers (nbut1375) calculates n/1 + n/2 + n/3 + n/4.

[1, 1375] Guess The Number
Time Limit: 1000 MS memory limit: 65535 K

Problem description
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
Prompt
No source
ZOJ 3175 Number of Containers

 

 

Question:

Calculate n * (1/1 + 1/2 + 1/3 + ...... 1/(n-2) + 1/(n-1)-n
Value

Because n is very large, it cannot be directly violent.

 

 

The number of All Integer Points on the line is n/I.

So n/1 + n/2 + n/3 + ...... N/(n-2) + n/(n-1) + n/n can be expressed as I * (n/I) = n this line

The answer is the number of Integer Points in the area enclosed by the line and coordinate axis.

Draw a line x = y and x * y = n to know the area is symmetric with x = y.

Let's calculate n/1 + n/2 + n/3 + ...... Obtain k = sqrt (n) (1 trapezoid) and multiply it by 2 (the area of 2 trapezoid is duplicated) subtract k * k duplicate regions

 

[Cpp]
# Include <stdio. h>
# Include <math. h>

Int main ()
{
Int t;
Scanf ("% d", & t );
Int n;
While (t --)
{
Int I;
Int t;
Long sum = 0;
Scanf ("% d", & n );
T = (int) sqrt (double) n );
For (I = 1; I <= t; I ++)
Sum + = (n/I );
Sum * = 2;
Sum = sum-t * t-n;
Printf ("% lld \ n", sum );
}
Return 0;
}

# Include <stdio. h>
# Include <math. h>
 
Int main ()
{
Int t;
Scanf ("% d", & t );
Int n;
While (t --)
{
Int I;
Int t;
Long sum = 0;
Scanf ("% d", & n );
T = (int) sqrt (double) n );
For (I = 1; I <= t; I ++)
Sum + = (n/I );
Sum * = 2;
Sum = sum-t * t-n;
Printf ("% lld \ n", sum );
}
Return 0;
}
In addition, we can use this method to quickly obtain n/1 + n/2 + n/3 +... n/n