There are two labels in one string!

Почему в тесте 1 ответ 2.

is there only one TALLES soldier or there can me more of them

hello .this is my programm
#include<stdio.h>
//#include<conio.h>
#include<math.h>
//
long energy,armor;
long kol_vrag,sr_energy_vrag,ugol;
long kol_robot,sr_energy_robot;
long ugol_otklona;
//------------------------
void Input(void);
void DG(void);
void DD(void);
void DA(void);
void DP(void);
int Min(const long x,const long y);
void Back(void);
void Atak(void);
//------------------------
void main(void)
{
 //clrscr();
 //
 Input();
 //
 //getch();
}
//------------------------
void Input(void)
{
 char vid;
 //
 scanf("%ld",&energy);
 scanf("%ld",&armor);
 //
 scanf("%c",&vid);
 scanf("%c",&vid);
 scanf("%ld",&kol_vrag);
 scanf("%ld",&sr_energy_vrag);
 scanf("%ld",&ugol);
 //
 switch(vid){
   case 'G':DG();break;
   case 'D':DD();break;
   case 'A':DA();break;
   case 'P':DP();break;
 }
}
//
void DG(void)
{
 if(!kol_vrag)
   printf("STOP");
 else
    if(abs(ugol)<5)
      printf("FIRE %d",Min(20,armor));
    else
    {
     if(ugol>=5)
       printf("LEFT %d",Min(100,energy));
     else
         if(ugol<=(-5))
           printf("RIGHT %d",Min(100,energy));
    }
}
//
void DD(void)
{
 if((20*kol_vrag)>=armor)
   Back();
 else
     DG();
}
//
void DA(void)
{
 scanf("%ld",&kol_robot);
 scanf("%ld",&sr_energy_robot);
 //
 if((kol_robot*sr_energy_robot)>(kol_vrag*sr_energy_vrag*3))
   Atak();
 else
     Back();
}
//
void DP(void)
{
 scanf("%ld",&ugol_otklona);
 //
 if(kol_vrag)
   DD();
 else
     if((abs(ugol_otklona)>20)&&(abs(ugol_otklona)<160)){
       if( ( (ugol_otklona>20)&&(ugol_otklona<=(90) ))||
      ((ugol_otklona<=-90)&&(ugol_otklona>-160) ) )
     printf("LEFT %d",Min(100,energy));
       else
       printf("RIGHT %d",Min(100,energy));
     }
     else
     printf("FRONT %d",Min(100,energy));
}
//
int Min(const long x,const long y)
{
 int r=x;
 //
 if(y<r)
   r=y;
 //
 return(r);
}
//
void Back(void)
{
 if(abs(ugol)<5)
   printf("FIRE %d",Min(20,armor));
 else
     printf("BACKWARD %d",Min(100,energy));
}
//
void Atak(void)
{
 if(abs(ugol)<10)
   printf("FIRE %d",Min(20,armor));
 else
     printf("FRONT %d",Min(100,energy));
}

Edited by author 06.02.2007 05:29

The problem statement says that a word may be followed by the end of file. So it is possible that the last line of a test is not followed by "end of line" character (or character sequence).

I know AC solution that outputs end of line after each line of output. Its behaviour contradicts to phrase in the statement '"reversing" the text twice restores the original text'.

I think that text with end of line before end of file and text without end of line before end of file are two different ones. So solution inserting extra end of line (when last line does not have one) should not get AC.

Admins! Please cut AC solutions off the following pages:

http://acm.timus.ru/forum/thread.aspx?space=1&num=1226&id=7841&upd=631727928559893810

http://acm.timus.ru/forum/thread.aspx?space=1&num=1226&id=9232&upd=631954522680635479

http://acm.timus.ru/forum/thread.aspx?space=1&num=1226&id=10920&upd=632452286387656250

http://acm.timus.ru/forum/thread.aspx?space=1&num=1226&id=12123&upd=632908242029782500

Извиняюсь, что на русском, но английский знаю плохо. В условии задачи на русском языке ошибка - "Тёмыч выгуливает своего любимого козла..." и "...число R — длину веревки, на которую Вася привяжет козла..."

#include <stdio.h>
#pragma warning (disable:4996)
#define min(a, b) ((a)<(b)?(a):(b))

int d[1001];

int main()
{
    int n, i, j;
    scanf("%d", &n);
    d[1] = 0;
    d[2] = 1;
    for (i =  3; i <= n; i++)
    {
        d[i] = 999999;
        for (j = 1; j < i; j++)
            d[i] = min(d[i],d[j] + d[i-j] + 1);
        for (j = 2; j < i; j++)
            if (i % j == 0)
                d[i] = min(d[i], d[i/j] + d[j]);
    }
    printf("%d\n", d[n]);
    return 0;
}

#include <stdio.h>
#pragma warning (disable:4996)

int a[1000];
int d[1000];

int main()
{
    int n, i, j, ans;
    scanf("%d", &n);
    for (i = 0; i < n; i++)
        scanf("%d", a+i);
    d[0] = 1;
    for (i = 1; i < n; i++)
    {
        d[i] = 1;
        for (j = 0; j < n; j++)
            if (a[j] < a[i])
                if (d[j] + 1 > d[i])
                    d[i] = d[j]+1;
    }
    ans = 1;
    for (i = 0; i < n; i++)
        if (d[i] > ans)
            ans = d[i];
    printf("%d\n", ans);
    return 0;
}

I got too many wa on it..

The subject test data is too strange

I input:
6
1 10
1 12
2 14
2 23
5 17
5 18
output:
4
3 4 6 5  test is wa#15

I change algorithm ， input again

output
4
3 4 5 6    test is wa #9

but I think both the result are all right, why?



Edited by author 23.12.2008 05:33

I have WA#4.

Dear Admins, don't you think that it would be prudent to state(or at least give a little hint), that WHITE IS TO MOVE FIRST?!!! Or is it so self-evident?!
Thank you.

I got Time limit exceeded in test 17. Can you tell me the reason ???

The numbers in input files (in tests) are not integer, but real!
But all numbers are in range 1..2^31 - 1

My prog passes all 307 tests from the forum and gets WA6, help me please... )

var
c:array [1..101] of integer;
i,k,j,kol,d:integer;
begin
kol:=0;
read(k);
for i:=1 to k do
 read(c[i]);
for j:= k-1 downto 2 do
 for i:= 1 to j do
  if (c[i]>c[i+1]) then
   begin
    d:=c[i];
    c[i]:=c[i+1];
    c[i+1]:=d;
   end;
    k:=round(k/2);
  for i:=1 to k do
    kol:=kol+round(c[i]/2);
  write(kol);
end.