Hướng dẫn giải của Bật đèn


Chỉ dùng lời giải này khi không có ý tưởng, và đừng copy-paste code từ lời giải này. Hãy tôn trọng người ra đề và người viết lời giải.
Nộp một lời giải chính thức trước khi tự giải là một hành động có thể bị ban.

Lưu ý: Các code mẫu dưới đây chỉ mang tính tham khảo và có thể không AC được bài tập này

Code mẫu của flashmt

uses math;
const maxn=100010;
      fi='';
var n,m,x,y,z,i,re:longint;
    a:array[1..maxn*5] of longint;
    b:array[1..maxn*5] of byte;

procedure add(node,l,r,x,y:longint);
var mid,t,u:longint;
begin
     if (l=x) and (r=y) then
     begin
          a[node]:=r-l+1-a[node];
          b[node]:=b[node] xor 1;
          exit;
     end;
     mid:=(l+r) shr 1;
     if x<=mid then add(node shl 1,l,mid,x,min(y,mid));
     if mid<y then add(node shl 1+1,mid+1,r,max(mid+1,x),y);
     a[node]:=a[node shl 1]+a[node shl 1+1];
     if b[node]=1 then a[node]:=r-l+1-a[node];
end;

procedure calc(node,l,r,x,y,z:longint;var re:longint);
var mid,t,u:longint;
begin
     if (l=x) and (r=y) then
     begin
          if z=0 then re:=a[node]
          else re:=r-l+1-a[node];
          exit;
     end;
     mid:=(l+r) shr 1; t:=0; u:=0;
     z:=z xor b[node];
     if x<=mid then calc(node shl 1,l,mid,x,min(y,mid),z,t);
     if mid<y then  calc(node shl 1+1,mid+1,r,max(mid+1,x),y,z,u);
     re:=t+u;
end;

begin
     assign(input,fi); reset(input);
     read(n,m);
     for i:=1 to m do
     begin
          read(z,x,y);
          if z=0 then add(1,1,n,x,y)
          else
          begin
               z:=0;
               calc(1,1,n,x,y,z,re);
               writeln(re);
          end;
     end;
     close(input);
end.

Code mẫu của happyboy99x

#include<bits/stdc++.h>
using namespace std;

const int N = 1e5;
int tree[4*N], com[4*N], n;

void update(int x, int y, int k = 0, int l = 0, int r = n) {
    if(r <= l || r <= x || y <= l) return;
    if(x <= l && r <= y) ++com[k], tree[k] = r-l-tree[k];
    else {
        update(x, y, 2*k+1, l, (l+r)/2);
        update(x, y, 2*k+2, (l+r)/2, r);
        tree[k] = tree[2*k+1] + tree[2*k+2];
        if(com[k] % 2 == 1) tree[k] = r-l-tree[k];
    }
}

int get(int x, int y, int k = 0, int l = 0, int r = n) {
    if(r <= l || r <= x || y <= l) return 0;
    if(x <= l && r <= y) return tree[k];
    int res = get(x, y, 2*k+1, l, (l+r)/2) + get(x, y, 2*k+2, (l+r)/2, r);
    return com[k] % 2 == 0 ? res : min(r, y) - max(l, x) - res;
}

int main() {
    int q; scanf("%d%d", &n, &q);
    while(q-- > 0) {
        int t, x, y; scanf("%d%d%d", &t, &x, &y);
        if(t == 0) update(x-1, y);
        else printf("%d\n", get(x-1, y));
    }
    return 0;
}

Code mẫu của ladpro98

program lites;
uses    math;
const   maxn=100005;
        fi='';
var     it,lazy:array[1..4*maxn] of longint;
        n,m,i,p,x,y:longint;
        inp:text;

procedure update(k,l,r,i,j:longint);
var     m:longint;
begin
        if lazy[k]=1 then
        begin
                it[k]:=(r-l+1)-it[k];
                if l<>r then
                begin
                        lazy[2*k]:=lazy[2*k] xor 1;
                        lazy[2*k+1]:=lazy[2*k+1] xor 1;
                end;
                lazy[k]:=0;
        end;
        if (l>r) or (r<i) or (j<l) then exit;
        if (i<=l) and (r<=j) then
        begin
                it[k]:=(r-l+1)-it[k];
                if l<>r then
                begin
                        lazy[2*k]:=lazy[2*k] xor 1;
                        lazy[2*k+1]:=lazy[2*k+1] xor 1;
                end;
                exit;
        end;
        m:=(l+r) shr 1;
        update(2*k,l,m,i,j);
        update(2*k+1,m+1,r,i,j);
        it[k]:=it[2*k]+it[2*k+1];
end;

function get(k,l,r,i,j:longint):longint;
var     p,q,m:longint;
begin
        if (l>r) or (r<i) or (j<l) then exit(0);
        if lazy[k]=1 then
        begin
                it[k]:=(r-l+1)-it[k];
                if l<>r then
                begin
                        lazy[2*k]:=lazy[2*k] xor 1;
                        lazy[2*k+1]:=lazy[2*k+1] xor 1;
                end;
                lazy[k]:=0;
        end;
        if (i<=l) and (r<=j) then exit(it[k]);
        m:=(l+r) shr 1;
        p:=get(2*k,l,m,i,j);
        q:=get(2*k+1,m+1,r,i,j);
        exit(p+q);
end;

begin
        assign(inp,fi);reset(inp);
        readln(inp,n,m);
        for i:=1 to m do
        begin
                readln(inp,p,x,y);
                if p=0 then update(1,1,n,x,y)
                else
                writeln(get(1,1,n,x,y));
        end;
end.

Code mẫu của RR

{$R+,Q+}
{$Mode objFPC}
uses math;
const
  FINP='';
  FOUT='';
  MAXN=100000;
  snode=524288;
var
  bat,tat,val:array[1..snode] of longint;
  n,q:longint;
  f1,f2:text;
procedure openF;
begin
  assign(f1,FINP); reset(f1);
  assign(f2,FOUT); rewrite(f2);
end;
procedure closeF;
begin
  close(f1); close(f2);
end;
procedure swap(var a,b:longint);
var
  temp:longint;
begin
  temp:=a; a:=b; b:=temp;
end;
procedure update(u,v:longint);
  procedure visit(i,l,r:longint);
  var
    mid:longint;
  begin
    if (v<l) or (r<u) then exit;
    if (u<=l) and (r<=v) then
      begin
        if val[i]=0 then swap(bat[i],tat[i])
        else val[i]:=0;
        val[i<<1]:=1-val[i<<1];
        val[i<<1+1]:=1-val[i<<1+1];
        exit;
      end;
    mid:=(l+r)>>1;
    if val[i<<1]=1 then
      begin
        val[i<<1]:=0;
        swap(bat[i<<1],tat[i<<1]);
        val[i<<2+0]:=1-val[i<<2+0];
        val[i<<2+1]:=1-val[i<<2+1];
      end;
    if val[i<<1+1]=1 then
      begin
        val[i<<1+1]:=0;
        swap(bat[i<<1+1],tat[i<<1+1]);
        val[i<<2+2]:=1-val[i<<2+2];
        val[i<<2+3]:=1-val[i<<2+3];
      end;
    visit(i<<1,l,mid);
    visit(i<<1+1,mid+1,r);
    bat[i]:=bat[i<<1]+bat[i<<1+1];
    tat[i]:=tat[i<<1]+tat[i<<1+1];
  end;
begin
  visit(1,1,n);
end;
function get(u,v:longint):longint;
var
  kq:longint;
  procedure visit(i,l,r:longint);
  var
    mid:longint;
  begin
    if (v<l) or (r<u) then exit;
    if (u<=l) and (r<=v) then
      begin
        if val[i]=1 then
          begin
            val[i]:=0;
            swap(bat[i],tat[i]);
            val[i<<1]:=1-val[i<<1];
            val[i<<1+1]:=1-val[i<<1+1];
          end;
        inc(kq,bat[i]);
        exit;
      end;
    mid:=(l+r)>>1;
    if val[i]=1 then
      begin
        val[i]:=0;
        swap(bat[i],tat[i]);
        val[i<<1+0]:=1-val[i<<1+0];
        val[i<<1+1]:=1-val[i<<1+1];
      end;
    visit(i<<1,l,mid);
    visit(i<<1+1,mid+1,r);
  end;
begin
  kq:=0;
  visit(1,1,n);
  get:=kq;
end;
procedure build(i,l,r:longint);
var
  mid:longint;
begin
  if l=r then
    begin
      bat[i]:=0;
      tat[i]:=1;
      exit;
    end;
  mid:=(l+r)>>1;
  build(i<<1,l,mid);
  build(i<<1+1,mid+1,r);
  bat[i]:=bat[i<<1]+bat[i<<1+1];
  tat[i]:=tat[i<<1]+tat[i<<1+1];
end;
procedure solve;
var
  u,v,yc:longint;
begin
  for q:=1 to q do
    begin
      read(f1,yc);
      if yc=0 then
        begin
          read(f1,u,v);
          update(u,v);
        end
      else //if yc=1 then
        begin
          read(f1,u,v);
          writeln(f2,get(u,v));
        end;
    end;
end;
begin
  openF;
  read(f1,n,q);
  build(1,1,n);
  solve;
  closeF;
end.

Code mẫu của hieult

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<cassert>
#include<ctime>
#include<algorithm>
#include<iterator>
#include<iostream>
#include<cctype>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<list>
//#include<conio.h>
#define ep 0.000001
#define maxn 100011
#define mod 1000000000
#define TR(c, it) for(typeof((c).begin()) it=(c).begin(); it!=(c).end(); it++)
#define fi first 
#define se second

double const PI=4*atan(1);
double const oo = 1e19;

using namespace std;

typedef pair<int, int> II;
typedef vector<int> VI;
typedef vector<II> VII;
typedef vector<VI> VVI;
typedef vector<VII> VVII;

struct tree{
    int number, change;
    tree(){};
    tree(int _number, int _change){ number = _number, change = _change;}
};

tree empty = tree(0,0);

int n, m, u, v, chiso;
tree f[maxn * 4];

void update(int l, int r, int i){

    if(u > r || v < l){
        if(f[i].change) f[i].number = r - l + 1 - f[i].number;
        f[i + i].change ^= f[i].change;
        f[i + i + 1].change ^= f[i].change;
        f[i].change = 0; 
    return;
    }
    if(u <= l && v >= r){ 
        f[i].change = 1 - f[i].change;
        if(f[i].change) f[i].number = r - l + 1 - f[i].number;
        f[i + i].change ^= f[i].change;
        f[i + i + 1].change ^= f[i].change;
        f[i].change = 0;
        return;
    }
    int mid = (l + r) / 2;
    f[i + i].change ^= f[i].change;
    f[i + i + 1].change ^= f[i].change;
    f[i].change = 0;
    update(l, mid, i + i);
    update(mid + 1, r, i + i + 1);
    f[i].number = f[i + i].number + f[i + i + 1].number;
}

int get(int l, int r,int i){
    if(u > r || v < l) return 0;
    if(u <= l && v >= r){
        if(f[i].change) return r - l + 1 - f[i].number;
        return f[i].number;
    }
    int mid = (l + r) / 2;
    f[i + i].change ^= f[i].change;
    f[i + i + 1].change ^= f[i].change;
    if(f[i].change) f[i].number = r - l + 1 - f[i].number;
    f[i].change = 0;
    return get(l, mid, i + i) + get(mid + 1, r, i + i + 1);
}

int main(){
   // freopen("input.in","r",stdin); 
   // freopen("output.out","w",stdout);
    scanf("%d %d",&n,&m);
    for(int i = 1; i < 4 * n; i++) {
        f[i] = empty;
    }
    //printf("%d.......\n",f[1].length);
    for(int i = 0; i < m; i++){
      //  printf("**%d\n",i);
        scanf("%d %d %d",&chiso,&u,&v);
        if(chiso == 0){ update(1, n, 1);}
        else printf("%d\n",get(1, n, 1));
    }
     // getch();
}

Code mẫu của ll931110

{$MODE DELPHI}
Program LITES2;
Const
  input  = '';
  output = '';
  maxn = 100000;
Type
  rec = record
    val: integer;
    swt,pre: boolean;
  end;
Var
  fi,fo: text;
  n,m,u,v: integer;
  a: array[1..8 * maxn] of rec;

Procedure openfile;
Begin
  Assign(fi, input);
    Reset(fi);

  Assign(fo, output);
    Rewrite(fo);
End;

Procedure update(i,low,high: integer);
Var
  mid,left,right: integer;
Begin
  If (v < low) or (high < u) then exit;
  If (u <= low) and (high <= v) then
    Begin
      a[i].swt:= not a[i].swt;
      exit;
    End;

  mid:= (low + high) div 2;
  update(2 * i,low,mid);
  update(2 * i + 1,mid + 1,high);

  left:= a[2 * i].val;
  If a[2 * i].swt then left:= (mid - low + 1) - left;

  right:= a[2 * i + 1].val;
  If a[2 * i + 1].swt then right:= (high - mid) - right;

  a[i].val:= left + right;
End;

Function calc(i,low,high: integer): integer;
Var
  mid: integer;
Begin
  If (v < low) or (high < u) then exit(0);
  If (u <= low) and (high <= v) then
    If not (a[i].swt xor a[i].pre) then exit(a[i].val)
                                   else exit(high - low + 1 - a[i].val);

  mid:= (low + high) div 2;
  a[2 * i].pre:= a[i].swt xor a[i].pre;
  a[2 * i + 1].pre:= a[2 * i].pre;
  calc:= calc(2 * i,low,mid) + calc(2 * i + 1,mid + 1,high);
End;

Procedure swap(var x,y: integer);
Var
  t: integer;
Begin
  t:= x;
  x:= y;
  y:= t;
End;

Procedure solve;
Var
  i,q: integer;
Begin
  For i:= 1 to 8 * maxn do
    Begin
      a[i].val:= 0;
      a[i].swt:= false;
    End;
  a[1].pre:= false;

  Readln(fi, n, m);
  For i:= 1 to m do
    Begin
      Readln(fi, q, u, v);
      If u > v then swap(u,v);
      If q = 0 then update(1,1,n) else writeln(fo, calc(1,1,n));
    End;
End;

Procedure closefile;
Begin
  Close(fo);
  Close(fi);
End;

Begin
  openfile;
  solve;
  closefile;
End.

Code mẫu của skyvn97

#include<cstdio>
#define MAX   100100
int t[5*MAX];
int c[5*MAX];
int m,n;
void init(void) {
    scanf("%d",&n);
    scanf("%d",&m); 
    int i;
    for (i=1;i<=4*n;i=i+1) {
        t[i]=0;
        c[i]=0;
    }
}
void update(int i,int l,int r,int u,int v) {
    //printf("Updating %d %d %d %d %d\n",i,l,r,u,v);
    if (l>v) return;
    if (r<u) return;
    if (l>r) return;
    if (u<=l && r<=v) {
        c[i]=1-c[i];
        t[i]=r-l+1-t[i];
        //printf("   t[%d]=%d c[%d]=%d\n",i,t[i],i,c[i]);
        return;
    }
    if (c[i]>0) c[2*i]=1-c[2*i];
    if (c[i]>0) c[2*i+1]=1-c[2*i+1];
    int m=(l+r)/2;
    if (c[i]>0) t[2*i]=m-l+1-t[2*i];
    if (c[i]>0) t[2*i+1]=r-m-t[2*i+1];  
    c[i]=0;
    //printf("   t[%d]=%d c[%d]=%d\n",2*i,t[2*i],2*i,c[2*i]);
    //printf("   t[%d]=%d c[%d]=%d\n",2*i+1,t[2*i+1],2*i+1,c[2*i+1]);
    update(2*i,l,m,u,v);
    update(2*i+1,m+1,r,u,v);
    t[i]=t[2*i]+t[2*i+1];
}
int sum(int i,int l,int r,int u,int v,int change) {
    //printf("Geting sum %d %d %d %d %d with %d\n",i,l,r,u,v,change);
    if (l>v) return (0);
    if (r<u) return (0);
    if (l>r) return (0);
    if (u<=l && r<=v) {
        if (change>0) return (r-l+1-t[i]);
        else return (t[i]);
    }
    if (c[i]>0) change=1-change;
    int m=(l+r)/2;
    int left=sum(2*i,l,m,u,v,change);
    int right=sum(2*i+1,m+1,r,u,v,change);
    return (left+right);
}
void process(void) {
    int i,s,e,T;
    //int j;
    for (i=1;i<=m;i=i+1) {
        scanf("%d",&T);
        scanf("%d",&s);
        scanf("%d",&e);
        if (T==0) {
            update(1,1,n,s,e);
            //printf("After update %d %d\n",s,e);
            //for (j=1;j<=7;j=j+1) printf("%d ",t[j]); printf("\n");
            //for (j=1;j<=7;j=j+1) printf("%d ",c[j]); printf("\n");
        }
        else printf("%d\n",sum(1,1,n,s,e,0));
    }   
}
int main(void) {
    //freopen("tmp.txt","r",stdin);
    init();
    process();
    return 0;
}

Code mẫu của khuc_tuan

#include <cstdio>
#include <cstring>
using namespace std;

const int MAXN = 100040;

int sum[MAXN * 4], flip[MAXN * 4];

void update(int n, int l, int r, int x, int y) {
    if(x<=l && r<=y) {
        flip[n] ^= 1;
        sum[n] = (r-l+1) - sum[n];
        return;
    }
    int m = (l+r) / 2;
    if(x<=m) update(2*n,l,m,x,y);
    if(m<y) update(2*n+1,m+1,r,x,y);
    sum[n] = sum[2*n] + sum[2*n+1];
    if(flip[n]) sum[n] = (r-l+1) - sum[n];
}

int getsum(int n, int l, int r, int x, int y, int f) {
    if(x<=l && r<=y) {
        if(f) return (r-l+1) - sum[n];
        else return sum[n];
    }
    int m = (l+r) / 2;
    f ^= flip[n];
    int res = 0;
    if(x<=m) res += getsum(2*n,l,m,x,y,f);
    if(m<y) res += getsum(2*n+1,m+1,r,x,y,f);
    f ^= flip[n];
    return res;
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i=0;i<m;++i) {
        int c, u, v;
        scanf("%d%d%d", &c, &u, &v);
        if(c==0) update( 1, 1, n, u, v);
        else printf("%d\n", getsum( 1, 1, n, u, v, 0));
    }
    return 0;
}

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