0 10
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
YES?

I am using recursion to solve the problem. But, I get WA8. I have debugged the code; but could not find any issues.

What is the test case?

Thanks,

HELP PLEASE!

Whether the barycenter of the polygon is must on the break-line ?
thanks ~

import java.io.*;

class Assert {
    static void check(boolean e) {
        if (!e) {
            throw new Error();
        }
    }
}

class Scanner {

    StreamTokenizer in;

    Scanner(InputStream is) {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(is)));
        in.resetSyntax();
        in.whitespaceChars(0, 32);
        in.wordChars(33, 255);
    }

    String next() {
        try {
            in.nextToken();
            Assert.check(in.ttype == in.TT_WORD);
            return in.sval;
        } catch (IOException e) {
            throw new Error(e);
        }
    }

    int nextInt() {
        return Integer.parseInt(next());
    }
}

public class Main {

    PrintWriter out;
    Scanner in;

    void solve() {
        int n = in.nextInt();
        int a = in.nextInt() + in.nextInt();
        int b;
        for (int i = 1; i < n; i++) {
            b = in.nextInt() + in.nextInt();
            out.print(a + b / 10);
            a = b % 10;
        }
        out.print(a);
    }

    void run() {
        in = new Scanner(System.in);
        out = new PrintWriter(System.out);
        try {
            solve();
        } finally {
            out.close();
        }
    }

    public static void main(String[] args) {
        new Main().run();
    }

}

Hi everyone!

I had WA#12 for quite a long time, even after I had tried all the tests listed here. I finally got AC after I checked the input for space characters (isspace). After that I hardcoded a check for ' ', '\n', '\r', '\t', '\v', '\f', '\a', '\b', (and removed the isspace) and it gave me WA. In the C standart isspace gives true only for space, newline (\n), carrige return (\r), horizontal tab (\t), vertical tab (\v), form feed (\f) AND others depending on the locale... I cannot understand what kind of symbols they have put in this test case because my program doesn't allow any whitespace at all (I handle each character one by one, and I look for perfect match for isdigit, '+', '-', '\0', 'e', 'E', so I just cant miss a space character) I suggest review of this test case and removing any fancy characters found in it.

If someone wants to see my solution email me iskren [dot] chernev [at] gmail [dot] com

Необходимо сложить площади двух колец и вычесть площадь их пересечения?

Isn't answer 40 on T1? trapeze with bases 10 and 4.

Edited by author 27.03.2011 03:35

For tasks which were solved by more than ~2000 authors (or something).
For example http://acm.timus.ru/rating.aspx?space=1&num=1017
Please help! :)

tle at test 22...

Sent on timus_support a couple of good tests, which my program that works 0.25 sec on timus, passes in about 1 minute on my PC

What's the answer for these tests?

#1
8
4 3 2 1 4 3 2 1
2
1 2 2
3 4 2

#2
8
4 3 2 1 4 3 2 1
2
3 4 2
1 2 2

thanks.

Of cource this problem can be solved using Aho Corasick algorithm. But there is another much more simple, but not determenistic solution. A hash function can be used to compare substrings of the text and the words. First we should find the hash function for all the words. Then we should evaluate the hash function for each substring of the text that has the same length as some word. And finaly comparing this this values of hash functions gives us the answer.

Can't understand why we need date starting from which person was the oldest.
Also can't understand why this solution gets WA3.
Can anyone help?

import java.util.GregorianCalendar;
import java.util.Scanner;

public class P1759 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = Integer.valueOf(scanner.nextLine());
        GregorianCalendar birth = (GregorianCalendar) GregorianCalendar.getInstance();
        GregorianCalendar death = (GregorianCalendar) GregorianCalendar.getInstance();
        long max = 0;
        int result = 0;
        long longestLiveDeath = 0;
        for (int i = 0; i < n; i++) {
            String[] dates = scanner.nextLine().split("\\s");

            set(birth, dates[0]);
            set(death, dates[2]);
            long liveTime = death.getTimeInMillis() - birth.getTimeInMillis();
            if (liveTime > max || (liveTime == max && death.getTimeInMillis() < longestLiveDeath)) {
                max = liveTime;
                longestLiveDeath = death.getTimeInMillis();
                result = i;
            }
        }
        System.out.println(result + 1);
    }

    private static void set(GregorianCalendar date, String sDate) {
        String[] split = sDate.split("\\.");
        date.set(Integer.valueOf(split[2]), Integer.valueOf(split[1]), Integer.valueOf(split[0]));
    }
}

Edited by author 24.03.2011 13:45

if the first caracter has a number less than 5 you should think on cyclic form.
Exemple: the first carater given is d..the numro of 'd' is 3 and 3-5<0...so you have to add 26 and of course d comes y(because 3-5+26=24 and y has 24)...
i whish i am clarified the point...good luck

var
   a:array[1..1000]of integer;
   b:array[1..1000]of integer;
   n,m,i,j:integer;
   s:string;
begin
readln(n);
for i:=1 to n do
readln(a[i]);
readln(m);
for j:=1 to m do
readln(b[j]);

s:='NO';
for i:=1 to n do
    for j:=1 to m do
    if a[i]+b[j]=10000 then s:='YES';
writeln(s);
end.

var w,g,h,l,t:real;
begin
g:=981;
read(l,h,w);
if l/2>h then begin write('butter'); halt end;
t:=sqrt(2*(h-l/2)/g);
w:=frac(w*t/60);
if ((w>=0) and (w<=1/4)) or ((w>=3/4) and (w<=1)) then write('butter') else write('bread');
end.