ACM: uva 11401 - Triangle Counti…

Triangle Counting

 

 

You are given n rods oflength 1, 2…, n. You have to pick any 3 of them &build a triangle. How many distinct triangles can you make? Notethat, two triangles will be considered different if they have atleast 1 pair of arms with different length.

 

Input

 

The input for each case will have only a single positiveinteger n (3<=n<=1000000).The end of input will be indicated by a casewith n<3. This case shouldnot be processed.

 

Output

 

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

 

SampleInput 

5

8

0

 

Output for SampleInput

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.

 

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAX 1000005

long longf[MAX];

int main()
{
    int n;
   f[2]=0;
    for(longlong i=3; i <= 1000000; i++)
       f[i]=f[i-1]+((i-1)*(i-2)/2-(i-1)+i/2)/2;

   while(scanf("%d", &n) != EOF)
    {
  if(n < 3)break;
       printf("%lld\n",f[n]);
    }
    return0;
}