UVA11401 Triangle Counting

最新推荐文章于 2021-05-09 20:34:16 发布

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本文介绍了一种计算使用长度从1到n的棒子可以构成的不同三角形数量的方法。通过枚举最长边并利用数学不等式简化计算过程，文章提供了一个高效的解决方案，并附带源代码实现。

统计用大小 1到n的那条边能组成多少个三角形。

利用y+z>x转化为不等式 x-y<z<x 枚举x。当x >3时一共有（x-1）(x-2)/2种方法。其中有y=z这种不合法的情况。（因为每条边只能用一次），所以要减去重复的（x-1）（x-2）/2

Triangle Counting

Time Limit: 1000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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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 withn<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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>

using namespace std;

#define MAXN 1000100

 long long f[MAXN];

int main()
{
    f[3]=0;
    for(long long i=4;i<=1000000;i++)
        f[i]=f[i-1]+((i-1)*(i-2)/2-(i-1)/2)/2;
    int n;
    while(~scanf("%d",&n)&&n>=3)
        printf("%lld\n",f[n]);
    return 0;
}