UVA 11401 Triangle Counting

Question：
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 (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
Sample Output
3
22
题意大意：输入一个数，求1~n能组成边长各不相等三角形的个数。
思路：令x为最大的边为x，z的取值范围（x-y,x）开区间。解得得到0，1，2，3，，，x-2得到和（x-1）（x-2）/2。但其中有y=z的三角形，取值从(x-1)/2~(x+x-1)/2,其中共有x-1项，但每两项的值是相同的，例如4/2==5/2,所以共有（n-1）/2项y==z,。但总的三角形都被算了两遍。所以sum[n]=sum[n-1]+((n-1)(n-2)/2-(n-1))/2

#include <iostream>#include <cstdio>using namespace std;typedef long long LL;const int Max=1000005;LL a[Max];int main(){    LL n,i;    a[3]=0;    for(i=4;i<Max;i++)        a[i]=a[i-1]+((i-2)*(i-1)/2-(i-1)/2)/2;    while(1)    {        scanf("%lld",&n);        if(n<3)            break;        printf("%lld\n",a[n]);    }    return 0;}

体会：灵活运用递推。