本文主要是介绍UVa 11401 Triangle Counting (组合计数)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
11401 - Triangle Counting
Time limit: 1.000 seconds
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=469&page=show_problem&problem=2396
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 Output for Sample Input
| 5 8 0 | 3 22 |
首先算出当最大边为x的符合题意的个数有多少，再用递推式累加即可。（因为题目要算的是边长不超过n的个数）
完整代码：
/*0.029s*/#include<cstdio>long long f[1000010];int main()
{f[3] = 0;for (long long x = 4; x <= 1000000; ++x)f[x] = f[x - 1] + ((x - 1) * (x - 2) / 2 - (x - 1) / 2) / 2;int n;while (scanf("%d", &n), n >= 3)printf("%lld\n", f[n]);return 0;
}
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