I can't understand what's wrong in my code, it write 51 instead of 50

#include <iostream>
using namespace std;

int main()
{
    unsigned short a, b, c;

    cin >> a >> b;

    c = (b - a) + 1;

    if(c % 2 == 0)
        cout << c / 2;
    else
        cout << (c / 2) + 1;  /// 1! 2 3! 4 5! 6 7! 8 9! 10 11! 12 13! 14 15!

    return 0;
}

#include <stdio.h>
#include <math.h>
main()
{
    int a=0;
    float b=0;
    while (!feof(stdin))
{
   scanf("%d", &a);
   b=sqrt(a);
   if (feof(stdin))
   printf("%.4f\n",b);

}
}


It works fine at my PC. But it keeps saying wrong answer?

//if 'double s[130196]' before 'int main()' - all is OK!!
//but if 'double s[130196]' in 'int main()' then Runtime error (stack overflow)!!!!!!!
//-------------------------------------------------
//AND MAGIC №2:
//any number after 130196 (for example 130196,130197...) in 'double s[130196]' - all is OK!!
//BUT numbers before 130196, for example 130195, 130194... then Runtime error (access violation)!!!
//------------
//on the internet people use 'double s[131072]' - it is 2^17, why???? if i can use 130196
//------------------
#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

double s[130196];// if 130195 or < then Runtime error (access violation)!!!
int main()
{
        //double s[130196]; - if in 'int main()' then Runtime error (stack overflow)
    int t = 0;
    while (cin>>s[t]){t++;}
    while(t>0){cout<<fixed<<setprecision(4)<<sqrt((double)s[--t])<<endl;}
    return 0;
}
//why why why ??))

Edited by author 14.07.2015 17:12

import java.util.Scanner;

public class _1370 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        int digitos = in.nextInt();
        int cliques = in.nextInt();
        String saida = "";
        String parte2 = "";

        for (int i = 1; i <= cliques; i++){
            saida += in.nextInt();
        }

        for (int i = cliques+1; i <= digitos; i ++){
            parte2 += in.nextInt();
        }

        parte2 += saida;

        System.out.println( parte2.substring(0,10) );
    }
}

Can anybody send me the fifth test of this problem? My program always crushes on it.
I think it's something wrong when input my data

int size;
struct point{
    int number;
    bool pr;
};
vector<point> matrix[1000];
int main(){
    int name1, name2;
    point c;
    char hg='0';
    vector<int> victims;
    cin >> size;
    while (hg != 'B'){
        cin >> hg;
        if (hg != 'B'){
            cin >> name2;
            name1 = atoi(&hg);
            c.pr = true;
            c.number = name2 - 1;
            matrix[name1 - 1].push_back(c);
            c.pr = false;
            c.number = name1 - 1;
            matrix[name2 - 1].push_back(c);
        }
    }
for (int i = 0; i < 4; i++)
        cin >> hg;
while (std::cin >> name1)
        victims.push_back(name1 - 1);

987654321 => white
123456789 => white
666 => black
999 => white
8 => black
13 => white
19 => white
5 => white

Edited by author 26.12.2015 22:50

what is test 6?

I can't find any error. My program solve all tests,my I got WA1!!!
My code:
 var
 Z:array[1..151000] of longint;
 A,C:array[1..151000] of char;
 N,L,R,i,j,k,ind,M,t:longint;
 ch:char;
 find:boolean;


function Min(x,y:longint):longint;
begin
 if(x>y) then x:=y;
 Min:=x;
end;

BEGIN
 N:=0;

 ch:='a';
 while(ord(ch)>50) do begin
  read(ch);
  if(ord(ch)>50) then begin
   inc(N);
   A[N]:=ch;
  end;
 end;

 A[N+1]:='+';
 ind:=N+1;
 inc(N);
 readln;

 ch:='a';
 while(ord(ch)>50) do begin
  read(ch);
  if(ord(ch)>50) then begin
   inc(N);
   A[N]:=ch;
  end;
 end;


 Z[1]:=N;
 R:=0;
 L:=0;

 for i:=2 to N do begin
  if(i<=r) then Z[i]:=min(R-i+1,Z[i-L+1])
  else Z[i]:=0;
  while(i+z[i]<=N)and(A[i+z[i]]=A[z[i]+1]) do inc(z[i]);
  if(i+z[i]-1>r) then begin
   L:=i;
   R:=i+z[i]-1;
  end;
 end;


 find:=true;
 i:=N;
 while(i>=ind+1) do begin
  k:=i;
  while(i>=ind+1)and(Z[i]<k-i+1) do dec(i);
  if(i<ind+1) then find:=false;
  dec(i);
 end;

 if(find) then begin
   writeln('No ');
   i:=N;
   M:=1;
   while(i>=ind+1) do begin
    k:=i;
    while(i>=ind+1)and(Z[i]<k-i+1) do dec(i);
    t:=M;
    for j:=k downto i do begin
     C[t]:=A[j];
     inc(t);
    end;
    C[t+1]:=' ';
    M:=t+2;
    dec(i);
   end;

   for i:=M-2 downto 1 do begin
    write(C[i]);
   end;
 end
 else begin
  write('Yes');
 end;
END.

DFS, record the state, if come into a state appeared before , ignore it.
DFS is very slow, can anybody give some hints to DP?

#include <iostream>
#include <cmath>
#include <algorithm>

using namespace std;

int function(int n)
{
    if (n == 1)
        return 1;
    else
        if (n == 0)
        return 0;
    else{

        int k = 999999999;
        for (int i = 1; i <= sqrt(n); i++)
        {
            int a = function(n - i*i);
            if (a < k )
                k = min(k, a);
        }
        return function(k) + 1;



    }


}

int main()
{
    int n;
    cin >> n;
    cout << function(n) << endl;

}

package timus;

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class p1167 {
    static int a[], totals[];
    static int h, s;
    static int min = Integer.MAX_VALUE;
    public static void main(String[] args) throws FileNotFoundException {
        InputStream is = System.in;
//        is = new FileInputStream(new File("A.txt"));
        FastScanner sc = new FastScanner(new InputStreamReader(is));
        h = sc.nextInt();
        s = sc.nextInt();
        int k[] = new int[h];
        for (int i=0; i<h; i++) {
            k[i] = sc.nextInt();
        }
        totals = new int[h+1];
        totals[1] = k[0];
        for (int i=2; i<h+1; i++) {
            totals[i]=totals[i-1]+k[i-1];
        }
        int maxcap = h-s+1;
        int unhappines[][] = new int[h+1][h+1];
        for (int i=1; i<=/*maxcap*/h; i++) {
            int up = Math.min(h, maxcap+i);
            for (int j=i; j<=up; j++) {
                int len = j-i+1;
                int ones = totals[j]-totals[i-1];
                unhappines[i][j] = (len-ones)*ones;
            }
        }
        int a[][] = new int[h+1][s+1];
        for (int i=1; i<=h; i++) {
            Arrays.fill(a[i], Integer.MAX_VALUE);
        }
        int maxes[] = new int[s+1];
        for (int i=1; i<=s; i++) {
            maxes[i] = maxcap+i-1;
        }
        for (int i=1; i<=maxcap; i++) {
            a[i][1] = unhappines[1][i];
        }
        for (int stable=2; stable<=s; stable++) {
            for (int i=stable; i<=maxes[stable]; i++) {
                for (int j=i; j<=maxes[stable]; j++) {
                    int prev = a[i-1][stable-1];
                    a[j][stable] = Math.min(a[j][stable], unhappines[i][j]+prev );
                }
            }
        }
        System.out.println(a[h][s]);
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(Reader in) {
            br = new BufferedReader(in);
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

}

I don't know why my O(N) solution is crashing on 1st test.

Code on C++11: http://ideone.com/BZIoKv

Maybe I don't know some disadvantages of C++? Or it's my algorithm error?

Use cin cout for c++

Edited by author 29.10.2013 13:59

Are you sure the limit is properly set? (test 16, possibly 6*10^5 integers at output)

Does it matte how many couples there exist? And, eventually, when a
group of 10 hobbits remains unsplit, do I have to split it? I mean,
it's possible for some hobbits o remain unsplitted..... what should I
do with them?

Edited by author 23.08.2015 14:38

18 32
5 11
9 2
17 16
16 7
2 12
13 2
16 4
5 17
4 13
14 17
17 17
11 3
11 15
9 3
13 10
3 8
16 13
7 16
10 11
11 13
14 8
7 4
16 10
1 11
15 2
2 12
13 12
1 14
3 14
8 6
14 1
18 1

ans:65

If you use newton's method to find a square root make sure that your division algorithm for 2 big integers is fast enough to handle quickly a lot of divisions that requires newtons method, otherwise you will get TL ... even with O(n*log(BASE)) time complexity division algorithm your going to get TL.
Faster division at O(n) is described in Knuth's The Art of Computer Sience 2 edition, page 298 or 299...

So in case if yo get TL with newton's method (which has got actually better time complexity than binary search root finding) just use binary search algo for root with BASE = 10^9 for you big integers, it needs only division by 2, *, +, and - operations needed for big integer.
If input contains 600 digits you will only need 600 / 9 array elements to store that number ... and with less divisions you will get AC!

take care ...

Edited by author 09.07.2015 18:17

Found the mistake :)

Edited by author 09.07.2015 01:22

I solved the problem by using a topological sorting.The vertices are segments.The arcs of the graph characterize nesting