UVA 11401

Description

 

Problem G
Triangle Counting

Input: Standard Input

Output: Standard Output

 

 

You are given n rods of length 1, 2…, n. You have to pick any 3 of them & 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 (3<=n<=1000000). The end of input will be indicated by a case withn<3. This case should not be processed.

Output

 

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

Sample Input                                                  Output for Sample Input

5 8 0

3 22

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Problemsetter: Mohammad Mahmudur Rahman

设最大边为x的三角形有c(x)个，另外两条边分别为y和z，根据三角形不等式有y+z>x，所以z的取值范围x-y<z<x。当y等于1时，无解；y=2时，只有一个解，y=3时，有两个解，直到y=x-1时有x-2个解。根据等差数列求和，一共有（x-1）（x-2）/2个解。当然，这里面还包含了y=z的情况，而且每个三角形算了两遍。所以要统计y=z的情况。所以一共有1/2*((x-1)*(x-2)/2-(x-1)/2)种情况。

原题要求f（n）。根据加法原理f(n)=c（1）+c（2）+......+c(n).可以写成地推形式f(n)=f(n-1)+c(n)。

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;

const int maxn = 1000010;
long long counts[maxn];

int main()
{
    counts[3] = 0;
    for(long long x = 4; x <= 1000000; x++)
    {
        counts[x] = counts[x-1] + (((x-1)*(x-2))/2-(x-1)/2)/2;
    }
    int n;
    while(cin>>n && n >= 3)
    {
        cout << counts[n] << endl;
    }
    return 0;
}