i am sure that my program is ok, but wa#4

if (a == b*q+p) and (0 <= p < |b|) then
{
  q == a / b
  p == a % b
}

is it TRUE?

Edited by author 13.02.2007 14:34

[code deleted]

Edited by moderator 20.11.2019 00:22

Pascal's type TDateTime (module SysUtils) allow decide this problem in 10 strings.

Let's solve problems which have the same number with your rank in authors ranklist. Milanin is lucky=)

why WA?

give me test 5 please

var n,i:longint;
begin
read(n);
if n<=0 then begin
n:=n*-1;
write(((n*(n+1))div 2)*-1+1);
end
else writeln(n*(n+1)div 2);
end.

Edited by author 13.04.2010 16:42

Edited by author 13.04.2010 16:43

Что здесь поменять чтоб количество циклов было меньше...

var n,x,i,j,t:integer;
a,b:array[1..250000]of char;
tempchar,ch:char;
label fin,step;
begin
readln(n);
x:=-1;
for i:=1 to n do read(a[i]);
readln;
for i:=1 to n do read(b[i]);
readln;
i:=1;
while i<=n  do begin
j:=1;
                 while  j<=n do begin
                                if j=1 then begin
                                tempchar:=a[1];
                                a[1]:=a[n];
                                            end else begin
                                            ch:=a[j];
                                            a[j]:=tempchar;
                                            tempchar:=ch;
                                            end;
                                            j:=j+1;
                                end;
                 for t:=1 to n do if a[t]=b[t] then x:=i else begin
                                                              x:=-1;goto step;
                                                              end;
                 if i=n then x:=0;
                 goto fin;
                 step:i:=i+1;
                 end;
fin:write(x);
end.

Please, help.

Thanks.

increase size of array.
15000 (not 14999)
Edited by author 25.07.2010 03:39

Edited by author 25.07.2010 03:40

Here some testcases and my AC program's answers.
May be these would be helpful for you. Good Luck.

10 NO
100 YES
1000 YES
8 YES
16 NO
32 YES
64 NO
125 YES
126 YES
1 NO
12345678 YES
17 YES
107 YES
267 YES
2992 YES
987456 YES
769 YES
896 NO
1536 NO
796 YES

#include<iostream>
using namespace std;
void K(int);
void main()
{
    int b[10],j=0,k=0;
    for(;j<10;j++)
    {
        cin>>b[j];
        if(b[j]==0) break;
    }
    for(;k<j;k++)
        K(b[k]);
}






void K(int n)
{
    if(n!=1)
    {
        n++;
        int *a = new int [n];
        a[0]=0;
        a[1]=1;
        int i=2;
        for(;i<n;i++)
        {
            if(i%2==0){a[i]=a[i/2];}
            else {a[i]=a[(i-1)/2]+a[(i-1)/2+1];}
        }

        if(n%2==0){cout<<a[n-1];}
        else {cout<<a[n-2];}

        cout<<endl;
    }
    else
    {
        cout<<0<<endl;
    }
}

Checker accepts any number of digits in mantissa, so it is possible to output just exact answer having all digits. Rounding to 19 digits or appending zeros to 19 digits is not required.

For answer calculated in java.math.BigDecimal the following simple statement works:
out.println(sum.toPlainString() + "e0");

The sample input is: 5  5  8  13  27  14
output is: 3  !!!!
But (5+5+13+14)=37
    (8+27)=35
so 37-35=2!!!!!!!!!!!
why is 3  ??

"When a faryuk is sprayed upon with the deodorant, its aroma strength decreases by 25 percent. When a faryuk is scented, its aroma strength increases by 10 percent. (In both cases, the aroma strength can become fractional)."

The question is about what is the base for calculations of 10% and 25% if we spray one faryuk several times: is it 10% and 25% of _initial_ aroma strength of a faryuk, or 10%/25% of aroma strength which we got after first aromatization?

Help me please! What's wrong with my program??




#include <iostream>
using namespace std;

#define ll unsigned long long


ll f(ll* a)
{
    swap(a[4], a[5]);

    for (int i=1; i<=3; i++)
        if (a[2*i-1]>a[2*i])
            swap(a[2*i-1], a[2*i]);

    for (int c=1; c<=3; c++)
        for (int i=1; i<3; i++)
            if (a[2*i-1]>a[2*i+1])
            {
                swap(a[2*i-1], a[2*i+1]);
                swap(a[2*i], a[2*i+2]);
            }

    return a[6]+10*a[5]+100*a[4]+1000*a[3]+10000*a[2]+100000*a[1];
}


ll a[100001][7], sol[800][100010], sum;

int main()
{
    ll n, k=0, sum;
    cin>>n;

    for (ll i=1; i<=n; i++)
    {
        for (int j=1; j<=6; j++)
            cin>>a[i][j];

        sum=f(a[i]);
        //cout<<"i="<<i<<" "<<sum<<endl;

        bool f=true;
        for (ll cnt=1; cnt<=k; cnt++)
            if (sol[cnt][100009]==sum)
            {
                f=false;
                ll tmp=++sol[cnt][0];
                sol[cnt][tmp]=i;
                break;
            }

        if (f)
        {
            ++k;
            sol[k][0]=1;
            sol[k][1]=i;
            sol[k][100009]=sum;
        }
    }


    cout<<k<<endl;
    for (ll i=1; i<=k; i++)
    {
        for (ll j=1; j<=sol[i][0]; j++)
            cout<<sol[i][j]<<" ";
        cout<<endl;
    }

}

Well, after all, I solved this problem, but it took a long time for me to find all mistakes in my calculations.

My idea was to generate F[N,K] in a O(log(N)*log(N)) time, which means an amount of numbers with N or less digits (in a certain base) with K one-digits. Then I wrote a function C(N, K), which finds an amount of numbers with K one-digits in [1, N] using F and deleting its first digit every time. Actually it's all about dynamic programming.

But I think there must be more clear and efficient solutions, than mine (O(log(N)*log(N)) for time and memory). If it does, please, would you be so kind to share it?

The text is incorrect:

"...to present it by the set of pairs of adjacent edges..." means not to transform the graph into a form in which it will be unequivocally presented by the set of pairs of its adjacent edges (it is rather difficult problem), but only to share (divide, separate) it into pairs of adjacent edges (it is easy).

i.g. you can divide graph [1-2-3-4-1] into [1-2] and [3-4]