UVA - 11401 - Triangle Counting （递推！）

最新推荐文章于 2019-12-03 19:14:17 发布

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最新推荐文章于 2019-12-03 19:14:17 发布

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分类专栏： UVA ~~~~~~~~~~数学文章标签： ACM uva algorithm 数学

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本文介绍了UVA-11401问题——三角形计数，该问题要求从长度为1到n的杆中选择三根构成不同的三角形，并计算可能构成的不同三角形的数量。提供了递推公式及AC代码实现。

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UVA - 11401

Triangle Counting

Time Limit: 1000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Problem G
Triangle Counting

Input: Standard Input

Output: Standard Output

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

Problemsetter: Mohammad Mahmudur Rahman

Source

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Mathematics :: Combinatorics :: Others, Easier
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Mathematics :: Combinatorics :: Other Combinatorics
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 2. Mathematics :: Counting :: Examples
Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 5. Mathematics :: Combinatorics
Root :: Prominent Problemsetters :: Mohammad Mahmudur Rahman

递推过去。。

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
using namespace std;

LL f[1000010];

int main()
{
	f[3] = 0;
	for(LL x = 4; x <= 1000000; x++)
		f[x] = f[x-1] + ( (x - 1) * (x - 2) / 2 - (x - 1) / 2 ) / 2;		//递推公式 
		
	int n;
	while(cin >> n)
	{
		if(n < 3) break;
		cout << f[n] << endl;
	}
	return 0;
}