i have WA #2
#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
    int n, m;
    cin >> n >> m;

    int win = 0;
    for(int i = 0; i < m; i++)
    {
        int a;
        cin >> a;
        n -= a;
        if(i % 2 == 0)
            win = 2;
        else win = 1;
    }

    cout << win << endl;

    return 0;
}

First of all. If you get WA#1, check how you scan input stream, cause they've put some trash in file.

I used the following idea. The whole number with base k divisible by ( k - 1 ), if sum( div( digit[ i ] ) ) is divisible by ( k - 1 ). Where digit[ i ] is k-based digit in input number ( char from input stream ); div( digit[ i ] ) is a remainder of division digit[ i ] * k by ( k - 1 ). Let see digit on i-th position, for example, it's A. So in decimal implementation it's equal to A * k ^ i.
A * k ^ i = A * k^i - A * k^(i-1) + A * k^(i-1) = A * k^(i-1) * ( k - 1 ) + A * k^(i-1)
A * k^(i-1) * ( k - 1 ) is divisible by ( k - 1 ), so we have to check only A * k^(i-1). If we repeat the same logic to this expression (i - 2) times, we see, that we have to check division A * k / ( k - 1 ).

I tried my best to explain the main idea, sorry if it can't be understood.

input1
sd5=#124'ashds'#12#154'sah';
input 2
sd5:=#123+#12;

I got WA1 and I don't understand why. Give me some tests please.

What does it mean: "clockwise order"?
How do you define clockwise order in notconvex polygon?
For example: in picture like a snail what roundabout way is turned in the clockwise direction and what is not?

Sorry for bad Enlish...

how could it have been passed??? and with 0.031

Can anyone give me some tricky tests?

Can I get some test inputs please?

const Nmax=111111;
var
 path:array[1..Nmax] of char;
 N,M,K,pic,d,dmax,i,L,now:longint;
 ch:char;
 ans1,ans2,ans3:int64;

begin
 readln(N,M,K);
 L:=0;
 read(ch);
 while(ch='u')or(ch='d')or(ch='r')or(ch='l')or(ch='f')or(ch='b') do begin
  inc(L);
  path[L]:=ch;
  read(ch);
 end;

 pic:=1;
 d:=1;
 dmax:=1;
 now:=1;
 for i:=1 to L do begin
  if(path[i]='l') then begin
   if(now=1) then begin
    inc(d);
    if(dmax<d) then dmax:=d;
   end
   else begin
    dec(now);
    if(now=1) then begin
     d:=1;
     if(dmax<d) then dmax:=d;
    end;
   end;
  end;
  if(path[i]='r') then begin
   if(now=N) then pic:=N
   else begin
    inc(now);
    d:=0;
    if(now>pic) then pic:=now;
   end;
  end;
 end;

 ans1:=1;
 i:=2;
 while(i<=N)and(pic<N) do begin
  if(i-1>=dmax) then inc(pic);
  if(i-1>=dmax)and(pic<=N) then inc(ans1);
  inc(i);
 end;

 pic:=1;
 d:=1;
 dmax:=1;
 now:=1;
 for i:=1 to L do begin
  if(path[i]='d') then begin
   if(now=1) then begin
    inc(d);
    if(dmax<d) then dmax:=d;
   end
   else begin
    dec(now);
    if(now=1) then begin
     d:=1;
     if(dmax<d) then dmax:=d;
    end;
   end;
  end;
  if(path[i]='u') then begin
   if(now=M) then pic:=M
   else begin
    d:=0;
    inc(now);
    if(now>pic) then pic:=now;
   end;
  end;
 end;

 ans2:=1;
 i:=2;
 while(i<=M)and(pic<M) do begin
  if(i-1>=dmax) then inc(pic);
  if(i-1>=dmax)and(pic<=M) then inc(ans2);
  inc(i);
 end;


 pic:=1;
 d:=1;
 dmax:=1;
 now:=1;
 for i:=1 to L do begin
  if(path[i]='b') then begin
   if(now=1) then begin
    inc(d);
    if(dmax<d) then dmax:=d;
   end
   else begin
    dec(now);
    if(now=1) then begin
     d:=1;
     if(dmax<d) then dmax:=d;
    end;
   end;
  end;
  if(path[i]='f') then begin
   if(now=K) then pic:=K
   else begin
    d:=0;
    inc(now);
    if(now>pic) then pic:=now;
   end;
  end;
 end;

 ans3:=1;
 i:=2;
 while(i<=K)and(pic<K) do begin
  if(i-1>=dmax) then inc(pic);
  if(i-1>=dmax)and(pic<=K) then inc(ans3);
  inc(i);
 end;

 write(ans1*ans2*ans3);
end.

Рассуждения такие:
1)как только достигнуто значение верхней границы,то все последующие шаги бессмысленны,т.к. пути совпадут;
2)пока первые i точек попадают в минимум у 1-ой точки (т.е. можно построить график зависимости номера шага от его значения,причем за пределы значений нельзя выходить и потому строим график в этом случае параллельным оси шага),их всех(кроме 1-ой точки) не засчитываем,все оставшиеся(если есть) и дают ответ;
3)всё перемножаем.

6
one 7 4 5
two 6 3 10
three 5 10 9
four 7 6 7
five 2 5 8
six 3 3 3

I think : one, two, three, four, five.

Edited by author 18.03.2009 20:44

My submission get AC when i use Visual C++ 2013 and get TLE using G++ 4.9 C++11... Why?

It is guaranteed that the most often appeared letter is the right answer. Only one letter is needed.

Пишу на Dev-C++ примет ли мой код? и почему-то пишет Compilation Eror. вот код:
# include <iostream>
# include <windows.h>

using namespace std;
int main()
{
    int a, b;
    cout << "Enter your numbers ";
    cin >> a >> b;
    cout << "Answer is " << a+b << endl;

    system("pause");
    return 0;
}

#include <iostream>
using namespace std;
int main()
{
    int a,b;
    cin >>a;
    cin >>b;
    cout<<a+b;
    return 1;
}

search wikipedia for FERMAT'S LAST THEOREM and you will come to know everything about this problem.(1349 Farm)

Edited by author 01.11.2011 15:24

const Nmax=1000000;
var
 s:array[0..Nmax-1] of char;
 z,ans,rans:array[0..Nmax-1] of longint;
 N,M,i,L,R,last,len:longint;
 ch:char;

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

procedure Init;
begin
 N:=-1;
 read(ch);
 while(ord(ch)>=ord('a'))and(ord(ch)<=ord('z')) do begin
  inc(N);
  s[N]:=ch;
  read(ch);
 end;
 readln;

 inc(N);
 M:=0;
 s[N]:='#';
 read(ch);
 while(ord(ch)>=ord('a'))and(ord(ch)<=ord('z')) do begin
  inc(N);
  inc(M);
  s[N]:=ch;
  read(ch);
 end;
end;

BEGIN
 Init;

 z[0]:=N;
 L:=0;
 R:=0;

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

 len:=0;
 last:=N;
 R:=0;
 i:=N;
 while(i>=N-M) do begin
  if(z[i]>=last-i+1) then begin
   inc(r);
   ans[r]:=i;
   rans[r]:=last;
   inc(len,last-i+1);
   last:=i-1;
  end;
  dec(i);
 end;

 if(len<M) then writeln('Yes')
 else begin
  writeln('No');
  for i:=r downto 1 do begin
   for l:=ans[i] to rans[i] do write(s[l]);
   write(' ');
  end;
 end;
END.

В чём баг хз.Уже сколько пытаюсь сдать-всегда WA3!
Решал так:s1 и s2-входные строки,создаю строку s1#s2,вычисляю z-функцию для s=s1#s2,иду с конца строки до символа # и пытаюсь "покрыть" "не покрытые" символы. Если всю строку покрыл,то ответ No,иначе Yes

Only one subarray sum, isn't it?

I came up with a solution that deals with the problems mentioned in the other topics, but still get wa on test 7 after I tried all tests that I could think of. Why is this? Does anybody know what test 7 is? Thanks in advance! :)

Edited by author 13.10.2015 22:03

1 1
0

1 50000
5130

50000 1
2

19 40320
4204

p.s. why time limitation is 2sec? it can be solved very fast using Lucas theorem.

Edited by author 16.12.2011 12:47

WA #3

#include <iostream>
#include <cmath>
#include <map>
#include <vector>
#include <cmath>
#include <stdio.h>
#include <iomanip>
#include <math.h>
using namespace std;

int main()
{
    const double PI = acos(-1.);
    int n;
    double r;
    cin >> n >> r;

    double x1, y1;
    cin >> x1 >> y1;

    if(n == 1)
    {
        cout << fixed << setprecision(2) << r << endl;
        return 0;
    }

    double vaultx, vaulty;
    vaultx = x1;
    vaulty = y1;

    double x2, y2;
    double d;
    double p = 0.0;
    for(int i = 0; i < n - 1; i++)
    {
        cin >> x2 >> y2;

        d = sqrt(pow(x2 - x1, 2) + pow(y2 - y1, 2));
        p += d;

        x1 = x2;
        y1 = y2;
    }

    p += sqrt(pow(x1 - vaultx, 2) + pow(y1 - vaulty, 2));

    p += 2 * PI * r;

    cout << fixed << setprecision(2) << p << endl;

    return 0;
}