UVA 11401 - Triangle Counting

2013-10-28 13:33 95人阅读评论(0) 收藏举报

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 易做图. How many distinct 易做图s can you make? Note that, two 易做图s 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 易做图s you can make.

Sample Input Output for Sample Input

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;  
}

补充：软件开发 , C++ ,