Thanks you very much,

Oh, I'm stupid.
I understood the problem later. The jeep can carry only M liters. So, we can take M liters in the beginning, go on distance X, leave Y liters there and return back. So, 2 * X + Y <= M. Next time, we can take these Y liters and actually we would be able to carry M + Y liters.

type ac=array[1..601]of longint;
var a,c,mi,lan,ch:ac;
b,d,e,f,g,h,i,j,k,l,m,n:longint;t:array[1..601]of char;
function times(a,c:ac;n:longint):ac;
var i,j,l:longint;
begin
 fillchar(ch,sizeof(ch),0);
 for i:=n downto 1 do begin
  l:=0;
  for j:=n downto 1 do
   if i+j-n>0 then begin
    ch[i+j-n]:=ch[i+j-n]+a[i]*c[j]+l;
    l:=ch[i+j-n] div 10;
    ch[i+j-n]:=ch[i+j-n] mod 10;
   end;
 end;
 times:=ch;
end;
function larger(a,c:ac;n:longint):boolean;
var i:longint;
begin
 for i:=1to n do
  if a[i]>c[i] then exit(true)
  else if a[i]<c[i] then exit(false);
 exit(false);
end;
procedure goal(g:longint);
var i,j:longint;
begin
 if g=602 then begin
  for i:=601-m+1 to 601 do
   write(c[i]);
  WRITELN;
  halt;
 end;
 fillchar(lan,sizeof(lan),0);
 if odd(n) then
  for j:=2to (g-(601-m))*2 do
   LAN[j]:=a[j+(601-n)-1]
 else
  for j:=1to (g-(601-m))*2 do
   LAN[j]:=a[j+(601-n)];
 for i:=1to 10 do begin
  if i<10 then begin
   c[g]:=i;
   fillchar(mi,sizeof(mi),0);
   for j:=1to g-(601-m) do
    mi[j+g-(601-m)]:=c[j+(601-m)];
   mi:=times(mi,mi,(g-(601-m))*2);
  end;
  if i=10 then begin
   c[g]:=9;
   goal(g+1);
  end
  else if larger(mi,lan,(g-(601-m))*2) then begin
   c[g]:=i-1;
   goal(g+1);
  end;
 end;
end;
begin
 n:=0;
 repeat
  inc(n);
  read(t[n]);
 until eoln;
 for i:=1to n do
  a[601-n+i]:=ord(T[I])-48;
 h:=0;
 for i:=601 downto 1 do begin
  a[i]:=a[i]*2+h;
  h:=a[i] div 10;
  a[i]:=a[i] mod 10;
 end;
 for i:=1to 601 do
  if a[i]<>0 then break;
 n:=602-i;
 m:=(n+1)div 2;
 goal(601-m+1);
end.

I think it's OK.
Does anybody know what's wrong with it?

I will mail my solution for you !
I accepted in 0.046s and used 337 KB

To sort is a quite good way to save time.

You haven't to write some code like
cout<<"Enter amounts of colors and pieces\n";

and read FAQ before using Online Judge

There is a post with this test on the board)

In the problem write T1<=T3 and you write "400 50 160" it is wrong. sorry my english is very bad

I have a one question.
What here does dichotomy?
It here is not necessary :)
Maybe you don't understand this task?
Try to read again ;)

7+1+5<>0+6+8!!!
  Why did the answer said it is true???

For you are a chinese
I will give you my algo
cpp_student@163.com

As many as you want.

My compiler says
"Error: Not enough actual parameters"

>> add[i]:=a[i]+c[i]+j;
>> fillchar(times,sizeof(times),0);

Shame on you!

Soon I will use Inverse program 40 Islands -1419(I have Ac)
as validator of my own
and my suspictions will be confirmed

WA#7 After fixed
#include<iostream>
#include<algorithm>
using namespace std;
struct note
{
       int value;
       int rank;
};
note used[500000]={0};
bool cmp(note a,note b)
{
     return (a.rank>b.rank);
}
int main (void)
{
    int N,j=0;
    cin>>N;
    for (int k=0;k<N;k++)
    {
        bool zepo=false;
        int f;
        cin>>f;
        for (int i=0;i<j;i++)
        if (used[i].value==f)
        {
                              used[i].rank++;
                              zepo=true;
        }
        if (!zepo)
        {
                  used[j].value=f;
                  used[j].rank=1;
        }
    }
    sort (used,used+j,cmp);
    cout<<used[0].value<<endl;
    //cin>>j;
    return 0;
}

I am a kind person
Do you want AC code??????