you must print n^2 and n.

You can solve it o(1) just precalc answers fir all test by brute forse, than put it to array.
Proffit

I did some tests. They imply that the program gives correct results, but the system says: "wrong answer".... Why?

screenshot: http://i077.radikal.ru/1102/59/2854393a1c43.jpg

Excuse me for my english

hi all;
here's a bit simpler algorithm than in the question:

if(x > 0 && y > 0) {

    for(i = 1; i <= x + y; ++i) {

        y = x * x + y;
        x = x * x + y;
        y = sqrt((double)x - y);
        x -= (2 * y * y);
    }

}
could anyone prove it to me how this scary algorithm is the one in O(1),
i mean the one that tests reminders and then swap (if necessary).
thank you.

Edited by author 17.02.2013 15:05

Edited by author 17.02.2013 15:05

#include <stdio.h>
#include<stdlib.h>

 int main()
{
 int arr[5],arr1[5],arr2[5],i,a=0,b=0,n,x=0,y=0;
 printf("Enter number");
 scanf("%d",&n);
 x=n+1;
 y=n-1;
 while(n>0)
 {
    for(i=5;i>=0;i--)
    {
      arr[i]=n%10;
      n=n/10;
    }
 }


 a=( arr[0]+arr[1]+arr[2]);
 b=(arr[3]+arr[4]+arr[5]);
 //  printf("%d%d",a,b);
   if((a-b==1)||(a-b==-1))
      {
           while(x>0)
         {
           for(i=5;i>=0;i--)
           {
               arr1[i]=x%10;
               x=x/10;
           }
         }
          while(y>0)
         {
           for(i=5;i>=0;i--)
           {
               arr2[i]=y%10;
               y=y/10;
           }
         }
       if ((arr1[0]+arr1[1]+arr1[2]==arr1[3]+arr1[4]+arr1[5])||(arr2[0]+arr2[1]+arr2[2]==arr2[3]+arr2[4]+arr2[5]))
       { printf("\nYes");}
        else
       { printf("\nNo");}
      }
      else
      {
          printf("No");
      }
    return 0;
}

i am doing (n%k)+(n/k) after putting restriction on k(k<=1000) and n(n>=1). but my answer is  said to be wrong.please help.

Edited by author 28.06.2013 19:30

Edited by author 28.06.2013 19:30

#include<iostream>
using namespace std;

short a[50001],b[50001];
int N,M,index_1_0,index_1_10,index_2_0,index_2_10;
bool f_0,f_10,t_0,t_10;
const int T=10000;

int main()
{
    cin>>N;
    for(int i=0;i<N;i++)
    {
        cin>>a[i];
        if(a[i]>=0&&!f_0)
        {
            index_1_0=i;
            f_0=true;
        }
        if(a[i]>=1000&&!f_10)
        {
            index_1_10=i;
            f_10=true;
        }
    }
    cin>>M;
    for(int i=0;i<M;i++)
    {
        cin>>b[i];
        if(b[i]<=0&&!t_0)
        {
            index_2_0=i;
            t_0=true;
        }
        if(b[i]<=1000&&!t_10)
        {
            index_2_10=i;
            f_10=true;
        }
    }
    for(int i=0;i<=index_1_0;i++)
    {
        for(int j=0;j<=index_2_10;j++)
        {
            if(a[i]+b[j]<1000)
            break;
            if(a[i]+b[j]==T)
            {
                cout<<"YES";
                return 0;
            }
        }
    }
    for(int i=index_1_0;i<=index_1_10;i++)
    {
        for(int j=index_2_10;j<=index_2_0;j++)
        {
            if(a[i]+b[j]<1000)
            break;
            if(a[i]+b[j]==T)
            {
                cout<<"YES";
                return 0;
            }
        }
    }
    for(int i=index_1_10;i<N;i++)
    {
        for(int j=index_2_0;j<M;j++)
        {
            if(a[i]+b[j]<1000)
            break;
            if(a[i]+b[j]==T)
            {
                cout<<"YES";
                return 0;
            }
        }
    }
    cout<<"NO";
    }

Edited by author 28.06.2013 11:46

Hi,
 Can someone please help with the type of test case 4.

Thanks.

Sorry, can anyone explain difference
between using i++ or ++i
in loop
for(int i=1;i<=n;++i)
or
for(int i=1;i<=n;i++)

i tried print "i" in loop, all the same

#include<stdio.h>
#include<stdio.h>

void plus(char *a,char *b,char *c)
{
    char *p1,*p2,*p3;
    p1=a+99;
    p2=b+99;
    p3=c+99;
    while(*p1!=0||*p2!=0)
    {
        *p3+=*p1+*p2-2*'0';
        if(*p3>'9')
        {
            *p3-=10;
            *(p3-1)+=1;
        }
        p1--;
        p2--;
        p3--;
    }
}
void multi(unsigned char *s,int k)
{
    unsigned char *p;
    p=s+99;
    while(*p!='0')
    {
        *p+=(*p-'0')*(k-1);
        p--;
    }
    p=s+99;
    while(*p!='0')
    {
        while(*p>'9')
        {
            *p-=10;
            *(p-1)+=1;
        }
        p--;
    }
}
void main()
{
    unsigned char a[101],b[101],c[101],d[101];
    int n,k,i,j;
    for(i=0;i<100;i++)
    {
        a[i]=b[i]=c[i]=d[i]='0';
    }
    scanf("%d %d",&n,&k);
    k-=1;
    a[100]='\0';
    b[100]='\0';
    c[100]='\0';
    a[99]=k+'0';
    b[99]=(k+k*k)%10+'0';
    b[98]=(k+k*k)/10+'0';
    for(i=1;i<n-1;i++)
    {
        strcpy(c,d);
        plus(a,b,c);
        multi(c,k);
        strcpy(a,b);
        strcpy(b,c);
    }

    i=0;
    while(*(b+i)=='0')
    {
        i++;
    }
    printf("%s",b+i);

}

If your code on pascal, don't use "CASE x oF"

it's got CE(Compilation error)

Consider for instance the first drawing : in my perception, the staircase has 4 steps because there are 4 horizontal levels. So the first step (ie the lower one) is made up with 4 blocks, the second with 3 blocks, the third one with 2 blocks and the fourth with 2 blocks again ; so the sizes of the steps are not in _strictly_ descending order (the sizes are 4 3 2 2).

So what does "size of a step" really mean? the number of blocks between two consecutive horizontal levels ?  the height from the ground of a given level? the height between two consécutive levels?

"descending order" : how do you count the steps? from right to left? from top to bottom?


The task is no more than finding the number of partitions of n into at least two distinct parts.

Edited by author 29.06.2013 04:27

here is my code. In my computer he got AC on test#1. But in timus that got WA#1

var
 a:array[0..500010] of longint;

procedure sort(l,r:longint);
var
 i,j,p,x:longint;
begin randomize;
 i:=l; j:=r; x:=a[random(r-l+1)+l];
 repeat
  while x>a[i] do inc(i);
  while x<a[j] do dec(j);
  if i<=j then
   begin
    p:=a[i];a[i]:=a[j];a[j]:=p;
    inc(i); dec(j);
   end;
 until i>j;
 if i<r then sort(i,r);
 if l<j then sort(l,j);
end;

var
 p,j,k,n,i,temp,have:longint;
begin
 readln(n);
 for i:=1 to n do
  readln(a[i]);
 sort(1,n);
 temp:=-1;
 for i:=1 to n do
  begin
   if a[i]=temp then begin inc(have); continue; end;
   if n mod 2=1 then k:=1 else k:=0;
   if have>=n div 2+k then begin writeln(temp); halt; end;
   have:=1;temp:=a[i];
  end;
 if n=1 then writeln(temp);
end.

Transposing means swapping two tombstones NOT reversing the order of all the stones between selected two.

1...n print(i)
(n+1)...n+m print(i*n)

I have WA on test 8. Can someone help. I tried every given test in the site but I can't find any mistakes.
This is my code:
//
#include<iostream>
#include<cmath>
#include<algorithm>
#include<iomanip>
using namespace std;
bool b[2005][2005];
long double a[2005][2005];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    long double diag=sqrt(2)*100;
    int n,m,k,k1,r;
    cin>>n>>m>>r;
    for(int i=0;i<r;i++)
    {
        cin>>k>>k1;
        b[k][k1]=1;
    }
    n++;
    m++;
    for(int i=2;i<=n;i++)a[1][i]=a[1][i-1]+100;
    for(int i=2;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            a[i][j]=a[i-1][j]+100;
            if(j!=1 && a[i][j]>a[i][j-1]+100)a[i][j]=a[i][j-1]+100;
            if(b[i-1][j-1] && a[i][j]>a[i-1][j-1]+diag+0.001)a[i][j]=a[i-1][j-1]+diag;
        }
    }
    cout<<fixed<<setprecision(0)<<a[n][m]<<"\n";
    return 0;
}

If you got WA #3, check case n=1 ;)