#include<stdio.h>

main()
{
int a, sum=0;
scanf("%d",&a);
if(a<0&&a>=-10000)
{
  int i;
  for(i=a;i<=1;i++)
  {
    sum=sum +i;

}
printf("%d",sum);
}

else if( a>=0&&a<=10000)
{
     int i;
     for(i=1;i<=a;i++)
     {
      sum=sum +i;
      }
     printf("%d",sum);

}

}

#include <stdio.h>

int fact(int n, int k){
    int res = 1;
    if (n == 0)
        return res;
    while (n > k) {
         res *= n;
         n -= k;
    }
    if (n % k)
        res *= (n % k);
    else
        res *= k;
    return res;
}

int main () {
    char sym;
    int n = 0;
    int symCount = 0;
    scanf("%d ", &n);
    while (scanf("%1[!]", &sym) == 1) {
            symCount++;
    }
    printf("%d\n", fact(n, symCount));
    return 0;
}

Answer for n is n*(n+1)/2*(n+2)

For example if n=2 then answer is 2*3/2*4=12

My program only read integer and write integer (ready formula). Why it works 0.015 sec?

Edited by author 11.06.2009 00:15

Let a be the number of objects.
Let b be the number of objects in a group.
Let A[a] is the number of permutations of a objects ( A[2] = 2, A[3] = 2*3, A[4] = 2*3*4...)

So:
f[a][b] is the number of groups with b objects.
You have the formula:
f[a][b] = f[a-1][b] * b + f[a-1][b-1];
And the result = f[n][1] * A[1] + f[n][2] * A[2] + ... + f[n][3] * A[3]

A group is formed by equal objects in a relation.
For example:
a = b < c = d (2 groups ab and cd)
a = b < c < d (3 groups ab, c and d)
...

I think when N is increased to 1000, everyone will solve it by DP or maths, not cheat :-)

solve it with c++

algo are:
dp[i,j]=dp[i,k]&&dp[k+1,j]

main code are:
for(length=0;length<n;length++)
    {
        for(i=0;i+length<n;i++)
        {
            if(find(i,i+length,maintree))
            {
                dp[i][i+length][0]=9999;
                dp[i][i+length][1]=1;
                save[i][i+length]=choose;
            }
            else
            {
                bestadd=100;bestj=-1;
                for(j=i+1;j<i+length;j++)
                if(dp[i][j][0]!=-1&&dp[j+1][i+length][0]!=-1)
                {
                    if(dp[i][j][1]+dp[j+1][i+length][1]<=bestadd)
                    {
                        bestadd=dp[i][j][1]+dp[j+1][i+length][1];
                        bestj=j;
                    }
                }
                dp[i][i+length][0]=bestj;
                dp[i][i+length][1]=bestadd;
            }
        }
    }

hope it will do you a favor

i try to find but i don't know where i'm wrong ??

Who know test 10, please tell me !!! Tks very much

Hello,
How can my program get AC even it returns wrong answer for the first test?

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;


public class SandrosBook {

    public static void main(String args[]) throws IOException{

        InputStreamReader isr = new InputStreamReader(System.in);
        BufferedReader br = new BufferedReader(isr);

        PrintWriter out=new PrintWriter(System.out);

        int[] a=new int[26];

        String s=br.readLine();
        int i=1;
        int start=0,end=0;
        start=s.charAt(0)-97;
        int max=0;




        while(i<s.length()){
            char c=s.charAt(i);
            a[(int)c-97]++;
            i++;
        }

        int k=0;
      for(int ii=0;ii<26;ii++){
           if(a[ii]>max)
           {
           max=a[ii];
           k=ii;
           }
        }

           k=k+97;

        System.out.println((char)k);
        out.flush();
    }

}

There is my code. My stress works fine. The size of output program is something near 4/5 * length bytes.
Using gets, all endl are written as '@'. There is an assert, that i haven't forgotten some symbols. Please, help!
http://pastebin.com/WnQvat6g

Apologies, since I started this thread.
Understood the problem got AC.
when I placed two books on shelf side by side I understood the problem.

Thanks to all memebers who posted in other threads.

Regards
Anupam

Edited by author 11.03.2012 02:41

Just use long arithmetics and calculate 36*pow(55,k-1);

Edited by author 06.04.2009 19:06

Does non-decreasing mean that is an increasing sequence of numbers?

Consider the following test:

7
Wayne 2 Bruce Dick
Bruce 1 Brown
Dick 1 Grayson
Brown 1 Gordon
Grayson 1 Gordon
Gordon 1 Stephanie
Stephanie 1 Wayne
Master Wayne is going to disguise as Batman and stop the Joker! I must prepare the Batmobile for his quest!

My solution fails to print Stephanie as a person who received the tweet about the earthquake. My solution uses a shortest path algorithm and that is why it takes into consideration only the shortest tweet reaching Gordon(the tweet sent by Grayson). So Gordon will try to send the tweet that he received(on the path Wayne Dick Grayson) preappended with "RT @Grayson: "(which is 13 characters long), totalling 141 characters. So the algorithm concludes that Stephanie will receive no tweet. But if we take the path Wayne, Bruce, Brown to Gordon, Gordon will receive a longer tweet than on the other path, but Stephanie will receive a tweet totalling 140 characters so she will actually be notified by a tweet too, although my algo fails to notice this.

I believe that there are great chances that other shortest path algos implemented by other users behave the same, so please try to consider improving the tests with a test similar with mine,

Regards,
Ciprian

Edited by author 15.10.2012 00:16

if you get WA 4 and you solve the problem from the end, than (maybe) you must take into account that sensor return only possitive number or null    ans   intermediate value can be >= 0.

Edited by author 15.10.2012 02:29

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException  {
        Scanner sc = new Scanner(System.in);
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        int n = sc.nextInt();
        String S[] = new String[n];
        for (int i = 0; i < n; i++) {
            S[i] = sc.next();
        }
        int m = sc.nextInt();
        String[] S2 = new String[m];
        for (int i = 0; i < m; i++) {
            S2[i] = bf.readLine();
        }


        int count = 0, tempcont = 0;
        boolean isFineded = false;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {

                StringTokenizer st = new StringTokenizer(S2[i]);
                count -= tempcont;
                tempcont = 0;
                while (st.hasMoreTokens()) {
                    tempcont++;
                    String temp = st.nextToken();
                    if (isExits(temp, S[j])) {
                        count++;
                        isFineded = true;
                        System.out.println(i + 1 + " " + count);
                    } else {
                        count++;
                    }
                }
            }
            tempcont = 0;
        }

        if (isFineded == false) {
            System.out.println("Passed");
        }

    }
    static boolean isExits(String src, String str) {
        boolean bol=false;
        if (str.length() > src.length()) {
            return false;
        } else if (str.equals(src)) {
            return true;
        } else {
            for (int i = 0; i <= src.length() - str.length(); i++) {
                if (str.equals(src.substring(i, i + str.length()))) {
                    bol=true;
                }
                }
            return bol;
        }

    }
}

var
n,i,k:byte;
x,d,y:integer;
m:array[0..100] of integer;
begin
read(n,x);
for i:=1 to n do
read(m[i]);
k:=0;
y:=n;

if x<0 then begin

for i:=1 to y do begin
if (m[i]>x) and (m[i]<0) then
begin writeln('Impossible'); k:=k+1;end
else
begin if m[i]<0 then begin y:=abs(x); end;
if m[i]>0 then d:=m[i];
end;
if m[i]<d then d:=m[i];end;
writeln(d+d+abs(x));
writeln(y);

end;

if x>0 then begin
for i:=1 to n do begin
if (m[i]<x) and (m[i]>0) then begin writeln('Impossible'); k:=k+1;end
else
begin if (m[i]>x) and (m[i]>0) then begin d:=x; writeln(d);end;
if m[i]<0 then d:=m[i];end;
if (m[i]<0) and (m[i]>d) then d:=m[i];
end;
writeln(abs(d)+abs(d)+x);
end;




end.

Sorry for Russian...
моё решение задачи... перевожу число в двоичную чистему исчисления.. считаю количество единичек.. если это количество четное, то выиграл второй, если нечетное, то первый... Почему такое решение проходит все тесты??? Оно же неверное.. например при N=5 моя прога выдаст, что выиграл второй, хотя на самом деле выигрывает первый, взяв 2 камня...

found my mistake


Edited by author 26.06.2012 01:56

I've got AC by stupid brute force, like if (i % 2) k++;.
But i think that there is a easier way to solve this task. Topic.