UVA 11401:Triangle Counting

Triangle Counting

 

 

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 with n<3. This case should not be processed.

 

Output

 

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

 

Sample Input 

5

8

0

 

Output for Sample Input

3

22

 

题意:  给出1~n的数, 求出能组合成三角形的三个数有多少组, 每组内的数都要不一样.

 

解题思路:

          1. 训练指南里面很详细, 设x是最大的边, 其余的为y,z; 根据三角形性质有: x＞y+z 和 x-y < z < x;

          2. 解有:  0+1+2+3+...+(x-2) = (x-1)*(x-2)/2; 但是结果里面有y == z的情况和每个三角形算了两次.

             即: 最大边长为x的三角形有: ( (x-1)*(x-2)/2 - (x-1) + x/2 ) / 2;

          3. 最终结果: f[i] = f[i-1] + 最大边长为i的三角形的个数.

#include<stdio.h>
typedef long long ll;
ll f[1000005];
int main()
{
    int n;
    for(ll i=3; i <= 1000005; i++)
        f[i]=f[i-1]+((i-1)*(i-2)/2-(i-1)+i/2)/2;
    while(~scanf("%d",&n)&&n>2)
    {
        printf("%lld\n",f[n]);
    }
    return 0;
}