SSL JudgeOnline 1198——求逆序对数

Description

先给出逆序对的定义,对于一个给定的数列{An},如果有iAj,则称(i,j)为一逆序对.
求给出一个数列,求出这个数列包含多少个逆序对?

Input

n (<=10000)
n个数

Output

输出逆序对数?

Sample Input

Sample Output

5
3 4 2 1 3

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思路一：我们用归并排序，判断如果a[i]>a[j]就ans:=ans+mid-i+1，将后面的数全部加起来

思路二：我们用两个循环1:for i:=1 to n-1 do
2:for j:=i+1 to n do
判断每一种情况，如果a[i]>a[j]就加一

---

代码如下：

var  a:array[1..100000]of longint;
     n,ans:longint;

procedure init;
var i,j:longint;
begin
  read(n);
  for i:=1 to n do read(a[i]);
end;

procedure move1(l,mid,r:longint);
var  l1,i,j:longint;
     t:array[1..100000]of longint;
begin
  i:=l; j:=mid+1; l1:=0;
  while (i<=mid)and(j<=r) do
    if a[i]>a[j] then
      begin
        ans:=ans+mid+1-i;
        inc(l1);
        t[l1]:=a[j];
        inc(j);
      end
    else
      begin
        inc(l1);
        t[l1]:=a[i];
        inc(i);
      end;
  for i:=i to mid do
    begin
      inc(l1);
      t[l1]:=a[i];
    end;
  for i:=j to r do
    begin
      inc(l1);
      t[l1]:=a[i];
    end;
  for i:=l to r do a[i]:=t[i-l+1];
end;

procedure move(l,r:longint);
begin
  if l>=r then exit;
  move(l,(l+r) div 2);
  move((l+r) div 2+1,r);
  move1(l,(l+r) div 2,r);
end;

var i:longint;
begin
  init;
  move(1,n);
  write(ans);
end.