NYOJ 982 Triangle Counting (mathematical problem)

Triangle Counting time limit: 1000 MS | memory limit: 65535 KB

Description

You are given n rods of length 1, 2 ..., N. You have to pick any 3 of them and build a triangle. How many distinct triangles can you make? Note that, two triangles will be considered different if they have at least 1 pair of arms with different length.

Input

The input for each case will have only a single positive integer n (1 <=n <= 1000000 ). the end of input will be indicated by a case with n <1. this case shoshould not be processed.

Output

For each test case, print the number of distinct triangles you can make.

Sample Input

580

Sample output

322

N line segments with a length of 1-n are given. How many methods can be used to retrieve 3 different line segments from the n line segments, so that they can form triangles with three sides of length.

Analysis: the first method is triplicate loop enumeration, but the time complexity is O (n ^ 3. Even algorithms with O (n ^ 2) time are difficult to handle, so mathematical analysis is required.

There are C (x) triangles with the maximum side length x, y and z For the other two sides, and y + z> x for the triangle inequality. Therefore, the z range is x-y <z <x.

According to this inequality, when y = 1, the X-1 <z <x, no solution; y = 2 there is a solution (z = x-1 ); y = 3 There are two solutions (z = X-1 or z = X-2 )...... Until y = X-1 has a X-2 solution. A total of 0 + 1 + 2 + ...... + (X-3) + (X-2) = (x-1) (X-2)/2 solutions.

But this is not the correct value of C (x), because the above solution contains the case of y = z, and each triangle is calculated twice. Therefore, we need to calculate the condition of y = z. The value of y starts from x/2 + 1 to X-1, a total of (x-1)-(x/2 + 1) + 1 = x/2-1, however, when x is an odd number, the value of y is x/2. If x is an even number, the value of y is x/2-1. So to avoid discussing the parity of x, we can write x/2-1 as (x-1)/2 without affecting the correct results. Deduct this decomposition, and then divide it by 2, that is, C (x) = (x-1) (X-2)/2-(x-1)/2)/2; the original question requires "the maximum side length cannot exceed n of the number of triangles F (n)", then F (n) = C (1) + C (2) +... + C (n ). The recursive formula is f (n) = F (n-1) + C (n ).

#include
     
      long long ans[1000005];int main(){    ans[1] = ans[2] = ans[3] = 0;    for(long long x = 4; x <= 1000000; x++)        ans[x] = ans[x-1] + ((x-1)*(x-2)/2 - (x-1)/2) / 2;    int n;    while(~scanf("%d",&n))    {        if(n < 1) break;        printf("%lld\n",ans[n]);    }    return 0;}