C++ :

#include <iostream>
#include <cstdio>
#define MAX_NUM 105
using namespace std;

int main()
{
  int high[MAX_NUM], left[MAX_NUM], right[MAX_NUM], n;
  scanf("%d", &n);
  for(int i = 0; i < n; i++)
    scanf("%d", &high[i]);
  for(int i = 0; i < n; i++)
  {
    left[i] = 1;
    for(int j = 0;j < i; j++)
    {
      if(high[j] <high[i] && left[j] > left[i] - 1)   // 找出左边最长递增子序列
        left[i] = left[j] + 1;
    }
  }
  for(int i = n-1; i >= 0; i--)
  {
    right[i] = 1;
    for(int j = n-1;j > i; j--)
    {
      if(high[j] < high[i] && right[j] > right[i] - 1)  // 找出右边最长递减子序列
        right[i] = right[j] + 1;
    }
  }
  int max = 0;
  for(int i = 0; i < n; i++)
  {
    if(max < left[i] + right[i] - 1)
      max = left[i] + right[i] - 1;  // 寻找最多合唱人数
  }
  printf("%d\n",n - max);  
  return 0;
}

Pascal :

var n,i,j,x:longint;p2,p1,a:array[1..1000]of longint;
begin
  readln(n);
  for i:=1 to n do read(a[i]);
  p1[1]:=1;
  for i:=2 to n do
    begin
      x:=0;
      for j:=1 to i-1 do
        if (a[j]<a[i])and(p1[j]>x) then x:=p1[j];
      p1[i]:=x+1;
    end;
  p2[n]:=1;
  for i:=n-1 downto 1 do
    begin
      x:=0;
      for j:=i+1 to n do
        if (a[j]<a[i])and(p2[j]>x) then x:=p2[j];
      p2[i]:=x+1;
    end;
  x:=0;
  for i:=1 to n do
    if p1[i]+p2[i]-1>x then x:=p1[i]+p2[i]-1;
  writeln(n-x);
end.