区间问题：两道线段树和树状数组的基础练习

万能的线段树和非常简单但是和好用的树状数组，早有耳闻，但一直找不到题目练习，只是照着书上的程序打了一遍练练手而已。最近发现POJ和HDU的acm题库资源真的很丰富呃，先练练手复习一下吧

POJ 2352  求点左下区域的点数，因为数据已经按从下往上，从左到右排序，所以可以用统计当前1到x的点数和就是答案，树状数组即可解决。但是因为树状数组不能处理0下标，代码有点小改动

const
  maxn=100000;
var
  c,level:array[0..1000000] of longint;
  i,j,k,n,m,x,y:longint;
function lowbit(i:longint):longint;
begin
  exit(i and -i);
end;

procedure add(i,x:longint);
begin
  while i<=maxn do
  begin
    c[i]:=c[i]+x;
    i:=i+lowbit(i);
  end;
end;

function getsum(i:longint):longint;
begin
  getsum:=0;
  while i>0 do
  begin
    getsum:=getsum+c[i];
    i:=i-lowbit(i);
  end;
end;



begin
  readln(n);
  for i:=1 to n do
  begin
    readln(x,y);
    add(x+1,1);
    inc(level[getsum(x+1)-1]);
  end;
  for i:=0 to n-1 do
    writeln(level[i]);
end.

HDU 1754 没穿衣服的区间最值问题，线段树基础练习

我的代码仅能通过样例，提交显示编译失败，难道Delphi语法差那么多吗？我在FP下用Delphi模式也能编译通过啊……

type
  node=record l,r,c:longint; end;
var
  tree:array[1..20000] of node;
  a:array[1..20000] of longint;
  n,m,i,x,y:longint;
  maxnum:longint;
  c,c1:char;
function max(a,b:longint):longint;
begin
  if a>b then exit(a) else exit(b);
end;
procedure build(p,l,r:longint);
begin
  tree[p].l:=l;
  tree[p].r:=r;
  if l=r then
    tree[p].c:=a[l]
  else begin
    build(p*2,l,(l+r) div 2);
    build(p*2+1,(l+r) div 2+1,r);
    tree[p].c:=max(tree[p*2].c,tree[p*2+1].c);
  end;
end;
procedure find(p,l,r:longint);
var
  m:longint;
begin
   if (l<=tree[p].l)and(tree[p].r<=r) then
     maxnum:=max(maxnum,tree[p].c)
   else begin
     m:=(tree[p].l+tree[p].r) div 2;
     if r<=m then
       find(p*2,l,r)
     else if l>m then
       find(p*2+1,l,r)
     else begin
       find(p*2,l,m);
       find(p*2+1,m+1,r);
     end;
   end;
end;

procedure insert(p:longint);
var
  m:longint;
begin
  if tree[p].l=tree[p].r then
    tree[p].c:=y
  else begin
    m:=(tree[p].l+tree[p].r) div 2;
    if m>=x then
      insert(p*2)
    else insert(p*2+1);
    tree[p].c:=max(tree[p*2].c,tree[p*2+1].c);
  end;
end;

begin
  while not eof do
  begin
    readln(n,m);
    for i:=1 to n do
      read(a[i]);
    readln;
    build(1,1,n);
    for i:=1 to m do
    begin
      readln(c,x,y);
      if c='Q' then
      begin
        maxnum:=-1;
        find(1,x,y);
        writeln(maxnum);
      end
      else if c='U' then
        insert(1);
    end;
  end;
end.