please give me a test for WA4.

According to my current logic result should be (filled bucket weight - empty bucket weight) for each mathematician.

Edited by author 11.12.2013 19:12

How can gelu's army reach the castle in 11 days??

what is the limit for output size
it could be very big for n=10000 and s=100000 and if the coordinates is in center of hypercube then there is (s-1)*2^(n-1) by bishop and n*(s-1) by rock  squares that queen can move
how the hell should i output this??
or am i wrong???

plz help ASAP!!!

Edited by author 21.08.2008 07:36

Edited by author 21.08.2008 07:36

I read 3 chars and run the follow DFA:
                one on in/out put
lost:
empty:
in/out:
input/output:
put:

here is my code:

#include <iostream>
#include <cstdio>
#include <cassert>
#include <cstdlib>
#include <string>

using namespace std;

inline bool good(char C)
{
    return (C >= 'a' && C <= 'z');
}

int a[5][4] = {{0, 0, 0, 0},
               {1, 0, 2, 4},
               {1, 0, 2, 3},
               {1, 1, 2, 4},
               {0, 1, 0, 0}};

char c[3];

bool get_ans()
{
    int n_u = 0, pos = 1;

    if (!good(c[2])) {
        return (c[0] == 'i' && c[1] == 'n');
    }

    bool n2 = 0;
    while (true)
    {
        if (pos == 0) return 0;
        if (n2) return (pos != 0 && pos != 4);

        c[0] = c[n_u % 3];

        for (int i = 3 - n_u; i < 3; i++)
        {
            c[i] = getc(stdin);
            if (!good(c[i])) {
                if (i == 0) return (pos != 0 && pos != 4);
                if (i == 1) return 0;
                n2 = 1;
                break;
            }
        }


        if (!n2 && c[0] == 'o' && c[1] == 'n' && c[2] == 'e') {
            pos = a[pos][0];
            n_u = 3;
            continue;
        }

        if (c[0] == 'o' && c[1] == 'n') {
            pos = a[pos][1];
            n_u = 2;
            continue;
        }

        if (!n2 && c[0] == 'o' && c[1] == 'u' && c[2] == 't') {
            pos = a[pos][2];
            n_u = 3;
            continue;
        }

        if (c[0] == 'i' && c[1] == 'n') {
            pos = a[pos][2];
            n_u = 2;
            continue;
        }

        if (!n2 && c[0] == 'p' && c[1] == 'u' && c[2] == 't') {
            pos = a[pos][3];
            n_u = 3;
            continue;
        }
        return 0;
    }
}

int main()
{
//    assert(freopen("inp.in", "r", stdin) != NULL);

    int n; assert(scanf("%d", &n) == 1);

    c[0] = getc(stdin);

    for (int i = 0; i < n; i++)
    {
        bool ok = 0;
        c[0] = getc(stdin);

        if (good(c[0])) {
            c[1] = getc(stdin);
            if (good(c[1])) {
                c[2] = getc(stdin);
                ok = get_ans();
            } else {
                ok = 0;
            }
        }

        if (ok) printf("YES\n");
        else printf("NO\n");
    }
}

Edited by author 10.12.2013 21:13

Edited by author 10.12.2013 21:15

Edited by author 10.12.2013 21:22

If you have WA #4 try this simple test:
3 2
1 5
10
answer is of course
16
1 2

When I use G++11 I always get TLE,but once I try vc++2010 to submit my code , I got AC.
I wasted too much time.Any can expalin it for me ?
This is my code:#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
const int MAXN = 1000000 + 10;
int a[2 * MAXN];
using namespace std;
int main()
{
    int n;
    scanf("%d", &n);
    int i = 1;
    while (n--)
    {
        int t;
        scanf("%d", &t);
        if (t > 0)
            a[i++] = t;
        else if (t == -1)
            printf("%d\n", a[--i]);
        else if (t == 0)
        {
            if (i - 1 <= n)
            {
                int num = min(i - 1, n);
                for (int j = 1; j <= num; ++j)
                {
                    a[i] = a[i-num];
                    ++i;
                }
            }
        }
    }
    //system("pause");
    return 0;
}

Edited by author 26.04.2013 15:20

The following is my code.
First come a special case where n is 1.
If n is not 1, more work should be done.

First, the nails are sorted based on their x-coordination.
The leftmost nail is accessed, and compared to the second leftmost nail.
With some boundary case, all the nails can be sorted along the clockwise or counter-clockwise direction.

Finally, I measure the distances between adjacent nails with pythagorean theorm.
Together with 2*pi*r, which accounts for the round edges, should be the answer.
What exactly does the code fail Test #5?


import math

def nail(n, r):
    return 2 * math.pi * r

def pyth(a, b):
    return math.sqrt(a**2 + b**2)

def perimeter(coord):
    ans = 0.0
    for i in range(len(coord)):
        ans += pyth((coord[i][0] - coord[i-1][0]), (coord[i][1] - coord[i-1][1]))
    return ans

if __name__ == "__main__":
    n, r = raw_input().split()
    n = int(n)
    r = float(r)
    coord = []
    for each in range(n):
        i, j = raw_input().split()
        coord += [(float(i), float(j))]

    if n == 1:
        print "%.2f" % (2 * math.pi * r)

    else:
        coord.sort()
        directed_coord = []
        directed_coord += [coord.pop(0)]
        if coord[0][1] > directed_coord[0][1]:
            clockwise = True
            directed_coord += [coord.pop(0)]
        else:
            clockwise = False
            directed_coord += [coord.pop(0)]

        if clockwise == True:
            for i in coord[:]:
                if i[1] > directed_coord[-1][1]:
                    coord.remove(i)
                    directed_coord += [i]
            for i in coord[:]:
                if i[1] >= coord[-1][1]:
                    coord.remove(i)
                    directed_coord += [i]
            coord.reverse()
            directed_coord += coord
        else:
            for i in coord[:]:
                if i[1] < directed_coord[-1][1]:
                    coord.remove(i)
                    directed_coord += [i]
            for i in coord[:]:
                if i[1] <= coord[-1][1]:
                    coord.remove(i)
                    directed_coord += [i]
            coord.reverse()
            directed_coord += coord

        ans = nail(n, r) + perimeter(directed_coord)
        print "%.2f" % ans

This is one of the easiest problem in timus and I can't believe i'm getting WA1 for this problem. Im solving this problem using C++ and I'm using getline method from sstream to get the input. I'm storing the names in vectors and printing them with empty lines between the houses. I know Test 1 is usually the given data, but im getting WA1 even though I'm getting the same answer in my computer as given in the problem description. Does anyone know how to overcome this WA1 problem?

Edited by author 03.01.2012 15:53

My solution jumped from TL 19 to AC in under 0.3s when i replaced output via printf with writing each digit of a number via putchar. That just blew my mind.

Edited by author 09.12.2013 11:44

Could you tell me what is test N9?

O(k*(sqrt(n)/log(n) + log(n))).
hint: precalc primes table(so, we can factorize number using O(sqrt(N)/log(N)) operations).
gl.

Can somebody help me?
I don't know what to do. I tried a lot of tests and all answers were correct.
My code:
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<stdlib.h>
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
int main()
{
    long a, b, x, y, z, p, q;
    char ch;
    scanf("%d%d", &a, &b);
    scanf("%s", &ch);
    scanf("%d%d%d", &x, &y, &z);
    a=min(a, 100);
    if(ch=='A') scanf("%d%d", &p, &q);
    if(ch=='P') scanf("%d", &p);
    if(ch=='P') if(x>0) ch='D';
    else
    {
        if((abs(p)>20)&&(abs(p)<160)){
        if(((p>20)&&(p<=(90)))||((p<=-90)&&(p>-160)))
        printf("LEFT %d", a);
        else printf("RIGHT %d", a);
     }
     else if(abs(p)<=20) printf("FRONT %d", a);
     else printf("BACKWARD %d", a);
    }
    if(ch=='D') if((x>0)&&((x*20)>=b))
    if(abs(z)<5) printf("FIRE %d", min(20, b));
    else printf("BACKWARD %d", a);
    else ch='G';
    if(ch=='G') {
        if(x==0) printf("STOP");
        else {
            if(abs(z)<5) printf("FIRE %d", min(20, b));
            if(z>=5) printf("LEFT %d", a);
            if(z<=-5) printf("RIGHT %d", a);
        }
    }
    if(ch=='A') if((p*q)>(3*(x*y)))
    if((x>0)&&(abs(z)<10)) printf("FIRE %d", min(20, b));
    else printf("FRONT %d", a);
    else if((x>0)&&(abs(z)<5)) printf("FIRE %d", min(20, b));
    else printf("BACKWARD %d", a);
}

My code:

why WA#6

import java.io.*;
import java.util.*;

public class misol1607
{
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int a = in.nextInt();
        int b = in.nextInt();
        int c = in.nextInt();
        int d = in.nextInt();
        int r=c;
        while (a<c)
        {
            if ((a+b)>=c)
            {
                r=c;
                break;
            }
            else
            {
                a=a+b;
            }

            if (a>=(c-d))
            {
                r=a;
                break;
            }
            else
            {
                c=c-d;
            }
        }

        out.println(r);

        out.flush();
   }
}

double h[131072],a,s;
int main() {
    int i=0,j;
   freopen("input.txt", "rt", stdin);
    while(cin>>a){
    h[i]=a;
    s=sqrt(a);
    h[i]=s;
    i++;
    }
    for(int j=(i-1);j>=0;j--){
        cout<<setiosflags(ios::fixed)<<setprecision(4)<<h[j]<<endl;
    }
    return 0;
}

Can anyone explain me test #1;
London keeps 1st from day 12 to 21 (10 days) and 23 to 25 (3 days). Day 22th London = StPetersburg (=15b). Why output is London 14?

uses
  SysUtils;
var
  a : array[0..10000] of longint;
  n, m, i, x, y : longint;
begin
  readln(n);
  if n = 1 then
    writeln(1)
  else if n mod 4 = 0 then
    writeln(0)
  else if (n mod 4 = 3) or (n = 25) or (n = 5) then
    writeln(2)
  else
    writeln(1);
  readln;
end.

15
<>><<>
<<><<<
>><

answer is :
28.

my program is accepted.