UVA - 11401
| Time Limit: 1000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
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
|
5 8 0 |
3 22 |
Problemsetter: Mohammad Mahmudur Rahman
Source
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;
}
UVA - 11401 - Triangle Counting （递推！）
原文地址：http://blog.csdn.net/u014355480/article/details/43488317
