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

Code mẫu của happyboy99x

#include<cstdio>
#include<cmath>

/* F(x) = p[0] + p[1] * x + p[2] * x^2 + ... + p[d] * x^d */
int d; double p[12];
const double EPS = 1e-6;

double f(double x) {
    double res = 0, x2 = 1;
    for(int i = 0; i <= d; ++i) res += x2 * p[i], x2 *= x;
    return res;
}

int main() {
    scanf("%d", &d); for(int i = 0; i <= d; ++i) scanf("%lf", p+i);
    double l = -1e6, h = 1e6;
    if(f(l) > f(h)) for(int i = 0; i <= d; ++i) p[i] = -p[i];
    while(fabs(h - l) > EPS) {
        double x = (l + h) / 2, y = f(x);
        if(y > 0) h = x; else l = x;
    }
    printf("%d\n", int((l+h)/2*1000));
    return 0;
}

Code mẫu của RR

{$R+,Q+}
uses math;
const
  FINP='';
  FOUT='';
  eps=0.00001;
var
  x,temp,left,right,mid:double;
  a:array[0..11] of double;
  n:longint;
function cal(x:double):double;
var
  i:longint;
  kq:double;
begin
  kq:=0;
  for i:=n downto 0 do
    kq:=kq*x+a[i];
  exit(kq);
end;
begin
  assign(input,FINP); reset(input);
  assign(output,FOUT); rewrite(output);
  read(n);
  for n:=0 to n do read(a[n]);
  left:=-1000000; right:=-left; x:=cal(left);
  repeat
    mid:=(left+right)/2;
    temp:=cal(mid);
    if abs(temp)<eps then begin writeln(trunc(mid*1000)); exit; end
    else
      if temp*x<0 then right:=mid
      else left:=mid;
  until left>right;
  close(input); close(output);
end.

Code mẫu của hieult

#include <stdio.h>
//#include <conio.h>
#include <math.h>
long n;
double a[13];
double f(double x)
{
double T=0;
for(long i=0;i<=n;i++)
  T=T+a[i]*pow(x,i);
return T;
}
main()
{
double u[100],v[100];
long tu=0,tv=0;      
scanf("%ld",&n);
for(long i=0;i<=n;i++)
  scanf("%lf",&a[i]);
if(a[n]>0)
  {  
  u[0]=-1000000;
  v[0]=1000000;
  }
else
  {  
  u[0]=1000000;
  v[0]=-1000000;
  } 
while(v[tv]-u[tu]>0.00001||v[tv]-u[tu]<-0.00001)
  {
  //printf("%.0lf ",u[tu]);
  if(f((v[tv]+u[tu])/2)>0)
    {
    tv++;
    v[tv]=(v[tv-1]+u[tu])/2;
    }
  else
    {
    tu++;
    u[tu]=(v[tv]+u[tu-1])/2;
    }
  }
printf("%ld",long(u[tu]*1000));
//getch();
}

Code mẫu của ll931110

{$N+}
Program CRUELL2;
  Const
    input  = '';
    output = '';
      maxd = 500;
      maxv = 1000000;
       eps = 0.0001;
  Var
    a: array[0..maxd] of double;
    d: integer;

Procedure init;
  Var
    f: text;
    i: integer;
  Begin
    Assign(f, input);
      Reset(f);

    Readln(f, d);
    For i:= 0 to d do readln(f, a[i]);

    Close(f);
  End;

Function calc(x: double): double;
  Var
    i: integer;
    k: double;
  Begin
    k:= a[d];
    For i:= d - 1 downto 0 do k:= k * x + a[i];
    calc:= k;
  End;

Procedure solve;
  Var
    a,b,c: double;
    ya,yb,yc: double;
    f: text;
  Begin
    Assign(f, output);
      Rewrite(f);

    a:= -maxv;
    b:= maxv;

    While b - a >= eps do
      Begin
        ya:= calc(a);
        If ya = 0 then
          Begin
            Writeln(f, trunc(a * 1000));
            Close(f);
            exit;
          End;

        yb:= calc(b);
        If yb = 0 then
          Begin
            Writeln(f, trunc(b * 1000));
            Close(f);
            exit;
          End;

        c:= (a + b) / 2;
        yc:= calc(c);
        If yc = 0 then
          Begin
            Writeln(f, trunc(c * 1000));
            Close(f);
            exit;
          End;

        If ya * yc < 0 then b:= c else a:= c;
      End;

    c:= (a + b) / 2;
    Writeln(f, trunc(c * 1000));
    Close(f);
  End;

Begin
  init;
  solve;
End.