UVa 11401 Triangle Counting:组合计数

UVa 11401 Triangle Counting:组合计数2014-10-02 synapse7

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

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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
", f[n]）;return 0;}