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 |
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;
}
