Usaco Barn Repair solution report

Question:
On a dark, stormy night, the roof and doors of farmer John's cowshed were blown away. Fortunately, many cows are on vacation, so the cowshed is not full. The cowshed is lined up next to the other, and the ox stays in the room for the night. Some cowshed have cows, and some do not. All cowshed have the same width. Farmer John had to erect a new plank before the cowshed as soon as possible after the door was lost. His new timber supplier will supply him any length he wants, but the stingy supplier can only provide a limited number of boards. Farmer John wants to minimize the total length of the Board he purchased.
Given: the maximum number of boards that may be purchased is M (1 <= M <= 50); Total number of Cowshed S (1 <= S <= 200 ); total number of cattle in a cowshed C (1 <= C <= S); number of the cowshed where the ox is located stall_number (1 <= stall_number <= S ), calculate the minimum total length of the plank required to stop all cowshed with cattle. Output the minimum length of the required Board as the answer.
Question:
There are M wooden boards that indicate from the smallest number of Cowshed to the largest middle there is a M-1 segment can not be covered to find out every two cowshed with cattle interval how much distance, find the first M-1 in these distances, and these do not need to be covered, then the minimum total length of the board is the continuous length of the whole Cowshed (the smallest number to the maximum) minus the length of the M-1 does not need to be covered, the answer is displayed.
Code:
[Cpp]
/*
ID: lishicao
PROG: barn1
LANG: C ++
*/
# Include <iostream>
# Include <fstream>
# Include <cstring>
# Include <algorithm>
Using namespace std;
 
Ifstream fin ("barn1.in ");
Ofstream fout ("barn1.out ");
 
Bool cmp (int a, int B)
{
Return a> B;
}
 
Int main ()
{
Int length [205], num [205];
Int M, S, C, Count = 0, ans;
Int Max = 0, Min = 999999;
 
Memset (length, 0, sizeof (length ));
 
Fin> M> S> C;
For (int I = 0; I <C; I ++)
Fin> num [I];
 
Sort (num, num + C );
 
For (int I = 0; I <C; I ++)
{
If (num [I]> Max) Max = num [I];
If (num [I] <Min) Min = num [I];
If (I! = 0)
{
Length [Count] = num [I]-num [I-1]-1;
Count ++;
}
}
Ans = Max-Min + 1;
Sort (length, length + Count, cmp );
For (int I = 0; I <M-1; I ++)
Ans-= length [I];
Fout <ans <endl;
Return 0;
}