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

Code mẫu của RR

//Written by Nguyen Thanh Trung
{$R+,Q+}
{$Mode objFPC}
uses math;
const
  FINP          =       '';
  FOUT          =       '';
  MAXN          =       111;
var
  f1,f2         :       text;
  a,b,c,d       :       array[1..MAXN] of longint;
  m,n,time      :       longint;
  ln1,nn1       :       longint;
  ln2,nn2       :       longint;

procedure openF;
    begin
      assign(f1,FINP); reset(f1);
      assign(f2,FOUT); rewrite(f2);
    end;
procedure closeF;
    begin
      close(f1);
      close(f2);
    end;
procedure inp;
    var
      i:longint;
    begin
      read(f1,time);
      read(f1,n);
      for i:=1 to n do
        read(f1,a[i],b[i]);
      read(f1,m);
      for i:=1 to m do
        read(f1,c[i],d[i]);
    end;
procedure get1(time:longint);
    var
      i,u:longint;
    begin
      ln1:=0;
      for i:=1 to n do
        begin
          u:=time-a[i];
          if u>0 then
          if u mod b[i]=0 then ln1+=u div b[i]+1
          else ln1+=u div b[i]+1;
        end;
      time-=1; nn1:=1;
      for i:=1 to n do
        begin
          u:=time-a[i];
          if u>0 then
          if u mod b[i]=0 then nn1+=u div b[i]+1
          else nn1+=u div b[i]+1;
        end;
      if nn1>ln1 then nn1:=ln1;
    end;
procedure get2(time:longint);
    var
      i,u:longint;
    begin
      ln2:=0;
      for i:=1 to m do
        begin
          u:=time-c[i];
          if u>0 then
          if u mod d[i]=0 then ln2+=u div d[i]+1
          else ln2+=u div d[i]+1;
        end;
      time-=1; nn2:=1;
      for i:=1 to m do
        begin
          u:=time-c[i];
          if u>0 then
          if u mod d[i]=0 then nn2+=u div d[i]+1
          else nn2+=u div d[i]+1;
        end;
      if nn2>ln2 then nn2:=ln2;
    end;
procedure solve;
    var
      l,r,mid:longint;
    begin
      l:=0; r:=time;
      repeat
        mid:=(l+r)>>1;
        get1(mid);
        get2(time-mid);
        if (ln1>=nn2) and (ln2>=nn1) then
          begin
            writeln(f2,mid);
            exit;
          end
        else if (ln2<nn1) then r:=mid-1
        else l:=mid+1;
      until l>r;
      get1(r);
      get2(time-r);
      if nn2<ln2 then writeln(f2,l)
      else writeln(f2,r);
    end;

begin
  openF;
  inp;
  solve;
  closeF;
end.

Code mẫu của hieult

#include <stdio.h>
//#include <conio.h>



int main()
{
    //freopen("SVADA.in","r",stdin);
    long long T,n,m,a[101],b[101],c[101],d[101];
    scanf("%lld",&T);
    scanf("%lld",&n);
    for(int i = 1;i<=n;i++)
        scanf("%lld %lld",&a[i],&b[i]);
    scanf("%lld",&m);
    for(int i = 1;i<=m;i++)
        scanf("%lld %lld",&c[i],&d[i]);
    long long u = 0, v = T+1;
    while(v-u>1)
    {
         long long r = (u+v)/2;
         long long sq1 = 0,sq2 = 0;
         for(int i = 1;i<=n;i++)
         {
             if(r<a[i]);
             else 
                 sq1 = sq1+(r-a[i])/b[i]+1;
         }
         for(int i = 1;i<=m;i++)
         {
              if(T-r<c[i]);
              else
                  sq2 = sq2+(T-r-c[i])/d[i]+1;
         }
         if(sq2>sq1)
              u = r;
         else if(sq2<sq1) v = r;
         else { u = r;break;}
    }
    printf("%lld",u);
   // getch();
}

Code mẫu của ll931110

{$R+,Q+}
program SVADA;
const
  input  = '';
  output = '';
  maxn = 100;
var
  a,b,c,d: array[1..maxn] of longint;
  t,n,m: longint;
  res: longint;

procedure init;
var
  f: text;
  i: longint;
begin
  assign(f, input);
    reset(f);

  readln(f, t);

  readln(f, n);
  for i := 1 to n do readln(f, a[i], b[i]);

  readln(f, m);
  for i := 1 to m do readln(f, c[i], d[i]);

  close(f);
end;

function found(x: longint): boolean;
var
  k1,k2: longint;
  i,tmp: longint;
begin
  k1 := 0;
  for i := 1 to n do
    begin
      if x < a[i] then tmp := 0
        else tmp := (x - a[i]) div b[i] + 1;
      if tmp > 0 then k1 := k1 + tmp;
    end;

  k2 := 0;
  x := t - x;
  for i := 1 to m do
    begin
      if x < c[i] then tmp := 0
        else tmp := (x - c[i]) div d[i] + 1;
      if tmp > 0 then k2 := k2 + tmp;
    end;

  found := k1 > k2;
end;

procedure solve;
var
  inf,sup,med: longint;
begin
  inf := 1;
  sup := t;
  res := 0;

  repeat
    med := (inf + sup) div 2;
    if found(med) then sup := med else inf := med;
  until sup - inf <= 1;

  res := inf;
end;

procedure printresult;
var
  f: text;
begin
  assign(f, output);
    rewrite(f);
    writeln(f, res);
  close(f);
end;

begin
  init;
  solve;
  printresult;
end.

Code mẫu của khuc_tuan

#include <cmath>
#include <algorithm>
#include <cstdio>
#include <iostream>

using namespace std;

long long T;
int N, M;
int A[111], B[111], C[111], D[111]; 

long long get(int n, int *a, int *b, long long tg) {
    long long res = 0;
    for(int i=0;i<n;++i) {
        if(a[i]<=tg) {
            res += 1 + (tg-a[i]) / b[i];
        }
    }
    return res;
}

long long getTime(long long sodua, int n, int *a, int *b) {
    long long L = 0, R = T;
    while(L<R) {
        long long tg = (L+R) / 2;
        long long sl = get(n, a, b, tg);
        if(sl<sodua) L = tg + 1;
        else R = tg;
    }
    if(sodua<=get(n,a,b,L)) return L;
    else return -1;
}

int main() {
    cin >> T;
    cin >> N;
    for(int i=0;i<N;++i) cin >> A[i] >> B[i];
    cin >> M;
    for(int i=0;i<M;++i) cin >> C[i] >> D[i];

    long long L = 0, R = T * 201;

    while(L<R) {
        long long sodua = (L+R) / 2 + 1;
        long long t1 = getTime(sodua, N, A, B);
        long long t2 = getTime(sodua, M, C, D);
        if(t1==-1 || t2==-1 || t1+t2>T) R = sodua - 1;
        else L = sodua;
    }
    cout << getTime(L, N, A, B) << endl;
    return 0;
}