I just can't stop tle-ing.
Should I use 4-based?

Edited by author 11.03.2014 15:04

'The sample text that could be READED the same in both orders ArozaupalanalapuazorA' =)

Edited by author 10.03.2014 19:18

Edited by author 10.03.2014 19:18

can anybody provide a test case for this error

my method is quiet like a brute force but it seems i have forgotten sth

Edited by author 10.03.2014 18:44

The problem states that the root should be in the range of (0, n - 1), which looks like saying 0 < x < n - 1. However apparently x can be equal to n - 1! I spent several hours stuck on WA 2 just because of this. Please change it to [0, n - 1] or whatever is precise.

give me some test, please.

but O(n^2) algorithm

I got a problem with test #7 but I don't know which is this test.

Any know?

Edited by author 09.03.2014 01:14

Edited by author 09.03.2014 01:16

Edited by author 09.03.2014 01:16

I got a problem with test #1 but I don't know which is this test.

Any know?

Thank

According to my program, test #5 contains cycles. Isn't it forbidden by the statement?

Edited by author 18.01.2014 03:05

why for D==5 answer is "NO"?

zzza
zzz
answer:
a

It test help me. It's not test 3. Good luck. =)

#include <iostream>
using namespace std;
int    main()
{
    int a[3];
    int b[3];
    int sum,m1,m2;
    cin>>a[0]>>b[0];
    if(a[0]<0||b[0]<0||a[0]>10000||b[0]>10000) exit(1);
    sum=a[0]+b[0];

    cin>>a[1]>>b[1];
    if(a[1]<0||b[1]<0||a[1]>10000||b[1]>10000) exit(1);
    if(b[1]>b[0]||(a[1]+b[1]!=sum))exit(1);

    cin>>a[2]>>b[2];
    if(a[2]<0||b[2]<0||a[2]>10000||b[2]>10000) exit(1);
    if(a[2]>a[0]||(a[2]+b[2]!=sum))exit(1);
    m1=a[0]-a[2];
    m2=b[0]-b[1];

    cout<<m1<<' '<<m2;
    return 0;
}

Edited by author 05.03.2014 20:47

Fail at #14

Всем привет,
ищу статью Скиданова, в которой он рассказывает, как стал чемпионом Урала и завоевал золотые медали Чемпионата мира.
В списке публикаций года на snarknews, блогах Скиданова на codeforces и wordpress эту статью не нашел.
В статье рассказывается, как Александр забивал на учебу ради тренировок, есть фразы вроде: «Не забывайте, что вы – программисты, а не машины для решения задач», «Вам что, не приходилось препода первый раз видеть на экзамене».
Поделитесь ссылкой, кто помнит эту статью. Буду признателен.

WA 30. Can anyone help me? Or just give AC code? I have spent to much time on this problem.
email: LegionN7@i.ua

Update:
Accepted.

Edited by author 05.03.2014 18:52

Может, я чего и не понимаю, но почему подобный код получает TL#1?

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define inf 2100000000

int a[2][510],res[2][510];
char fy[110][510];
int n,m,min,i1,i2;

void Rec(int i, int j)
{
   if (i > 1) {
      if (fy[i][j] == j) Rec(i-1,j); else Rec(i,fy[i][j]);
   };
   printf("%d\n",j);
};

int main(void)
{
   int i,j;

   scanf("%d%d",&m,&n);

   for (j = 1; j <= n; j++) scanf("%d",&a[1][j]);
   for (i = 1; i <= n; i++) res[1][i] = a[1][i];
   for (i = 2; i <= m; i++)
      for (j = 1; j <= n; j++) res[i][j] = inf;

   for (i = 1; i < m; i++) {
      i1 = i%2; i2 = (i+1)%2;
      for (j = 1; j <= n; j++) scanf("%d",&a[i2][j]);
      for (j = 1; j <= n; j++) {
         res[i2][j] = res[i1][j] + a[i2][j]; fy[i+1][j] = j;
      };
      for (j = 2; j <= n; j++)
         if (res[i2][j] > res[i2][j-1] + a[i2][j]) {
            res[i2][j] = res[i2][j-1] + a[i2][j]; fy[i+1][j] = j-1;
         };
      for (j = n-1; j >= 1; j--)
         if (res[i2][j] > res[i2][j+1] + a[i2][j]) {
            res[i2][j] = res[i+1][j+1] + a[i2][j]; fy[i+1][j] = j+1;
         };
      };

   min = inf;
   for (i = 1; i <= n; i++)
      if (res[m%2][i] < min) {
         min = res[m%2][i]; i1 = i;
      };

   Rec(m,i1);
   return 0;
};

I don't know why my program has WA#1.
My computer gives correct answer.
Please help me, maybe you had the same problem.

t[i] can be 0