UVA 11401 - Triangle Counting

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

5 8 0

3 22

 

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Problemsetter: Mohammad Mahmudur Rahman

解题思路，分段法，f [n] 记录最长的那条边不超过n的放法数， 假设c[n] 为最长边为n的放法数  f [n] = f [n-1] + c[n] 。

假设最长边为z，其它为 x,y;

则   z-x < y < z

当  x=1 时， z-1<y<z  0 种方案，

当  x=2 时， z-2<y<z  1 种方案，

......................................................

当  x=z-1时， 1<y<z  z-2种方案

所以总计 (z-1)*(z-2)/2 种方案

但是其中包含了x与y想等的情况，因为 x取值为 [ z/2+1 , z-1 ] 区间内可能 x,y相等，共 (z-1)- (z/2+1)+1=z/2-1 种方案

而且x,y与 y,x认为为两个，重复计算了，所以除以2

所以 c[n]= [ (z-1)*(z-2)/2 -(z/2-1) ] / 2

#include <iostream>
using namespace std;

const int maxn=1000000;
unsigned long long f[maxn+10];
int n;

void ini(){
	f[3]=0;
	for(long long z=4;z<=maxn;z++){
		f[z]=f[z-1] + ( (z-1)*(z-2)/2 - (z/2 -1) )/2;
	}
}

int main(){
	ini();
	while(cin>>n && n>=3){
		cout<<f[n]<<endl;
	}
	return 0;
}