TOJ 3474.The Big Dance

题目链接：http://acm.tju.edu.cn/toj/showp3474.html

3474.   The Big Dance
Time Limit: 2.0 Seconds   Memory Limit: 65536K
Total Runs: 517   Accepted Runs: 343    Multiple test files

Bessie and the herd, N (1 ≤ N ≤ 2,200) conveniently numbered 1..N cows in all, have gone to a dance where plenty of bulls are available as dancing partners. This dance is known
as the "odd cow out" dance because of the way cows are chosen to dance with bulls.
The cows are lined up in numerical order and the 'middle' point is chosen. It either divides the line of cows exactly in half or it is chosen so that the first set of cows has just one more cow in it than
the second set. If exactly two cows are in the set, they are chosen to dance with bulls. If one cow is in the set, she is sent home with a consolation prize of a beautiful rose.
If the set has more than two cows in it, the process is repeated perhaps again and again until sets with just one or two cows emerge.
The two cow ID numbers are multiplied together and added to a global sum.
Given the number of cows at the dance, compute the global sum after all the eligible cows are chosen.
Consider a dance with 11 cows numbered 1..11. Here is the sequence of dividing them:

1     2     3     4     5     6  |  7     8     9     10     11

1     2     3  |  4     5     6

1     2  |  3
1  2        => 1*2=2 added to sum -> sum=2
3           => sent home with rose

4     5  |  6
4  5        => 4*5=20 added to sum -> sum=22
6           => sent home with rose

7     8     9  | 10    11

7     8  |  9
7  8        => 7*8=56 added to sum -> sum=78
9           => sent home with rose
10    11            => 10*11=110 added to sum -> sum=188

So the sum for this dance would be 188.

Input

* Line 1: A single integer: N

Output

* Line 1: A single integer that is the sum computed as prescribed.

Sample Input

11

Sample Output

188

简单的递归题，水题上代码：

#include <stdio.h>
int cow(int start, int end){
if(start == end)
return 0;
if(end == start + 1)
return end * start;
int middle = (start + end)/2;
return cow(start, middle) + cow(middle+1, end);
}
int main(){
int n;
while(~scanf("%d",&n))
printf("%d\n",cow(1, n));
return 0;
}