i see N%3 solution everywhere but does anyone have any explanation as to why the guy who has mod-3 no of stones should lose. Some logical or mathematical reason would be much appreciated.

why should i declare 128*1024 size of memory for 256KB described in the problem?
my code goes here:
#include<iostream>
#include <cstdio>
#include <cmath>

double buf[128 * 1024];
int main()
{
    int i = 0;
    unsigned long long n;
    while (scanf("%llu", &n) != EOF) {
        buf[i ++] = (double)sqrt(n * 1.0);
    }
    for (; --i >= 0; ) {
        printf("%.4lf\n", buf[i]);
    }
    return 0;
}

//  why 128*1024  ?

Hello,
I've not found a topic about this test. Does anyone know what is it?
I'm just searching from the half of the string to find where can be the center of a palindom.
Thank you

The follwing is my pgm. Could some give a test case not satisfied here?
var
s1,i:integer;
f:boolean;
a2,b2,a1,b1,a,b:real;
begin

 f:=false;
 read(a);
 read(b);
 a1:=a/100;
 b1:=b/100;
 if b<>0 then
 i:=trunc(1/b1);
 a2:=a1*(i);
 b2:=b1*(i);
 while not f do
    begin
   if (((trunc(b2)<>trunc(a2)))and (frac(b2)<>0)   then f:=true;
   a2:=a2+a1;
   b2:=b2+b1;
   i:=i+1;
 end;
 i:=i-1;
 writeln(i);

end.

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

const long long mod=1000000007;
const int nmax=200;

int n,m,t,p;
long long v[nmax],fact[nmax],step[nmax];
bool used[nmax];
bool a[nmax][nmax];
long long ans;

void DFS(int v)
{
       for(int j=1;j<=n;j++)
       {
              if(a[v][j] && !used[j])
              {
                     t++;
                     used[j]=true;
                     DFS(j);
              }
       }
}

bool FindCycle(int v,int pr)
{
     bool f=false;
     for(int j=1;j<=n;j++)
     {
            if(a[v][j] && j!=pr)
            {
                   if(used[j]) return true;
                   else
                   {
                     used[j]=true;
                     t++;
                     f=f | FindCycle(j,v);
                   }
            }
     }
     return f;
}

void main()
{
       scanf("%d%d\n",&n,&m);

        if(n<=2)
        {
               printf("1");
               return;
        }

       fact[0]=1;
       for(int i=1;i<=n;i++)
              fact[i]=fact[i-1]*((long long)i) % mod;

       for(int i=1;i<=n;i++) step[i]=0;

       for(int i=1;i<=n;i++)
       {
              for(int j=1;j<=n;j++)
                     a[i][j]=false;
       }

       for(int i=1;i<=m;i++)
       {
          scanf("%d\n",&t);
          if(!a[i][t])
          {
              a[i][t]=a[t][i]=true;
              step[i]++;
              step[t]++;

          }
       }

       p=1;
       while(p<=n && step[p]<=2) p++;
       if(p<=n)
       {
              printf("0");
              return;
       }

       for(int i=1;i<=n;i++) used[i]=false;

       for(int i=1;i<=n;i++)
       {
              if(!used[i])
              {
                   used[i]=true;
                   t=1;
                   if(FindCycle(i,-1))
                   {
                          if(t<n)
                          {
                            printf("0");
                            return;
                          }
                          if(t==n)
                          {
                            printf("2");
                            return;
                          }
                   }

              }
       }

       for(int i=1;i<=n;i++) used[i]=false;
       p=0;
       ans=1;
       for(int i=1;i<=n;i++)
       {
              if(!used[i])
              {
                     p++;
                     t=1;
                     used[i]=true;
                     DFS(i);
                     if(t>1) t=2;
                     ans*=((long long)t) % mod;
              }
       }
       ans=ans*fact[p-1] % mod;
       while(ans<0) ans+=mod;
       printf("%lld",ans);
}

a a b a a
answer: 1

If you have WA4 java. Put 10^7  instead of 10^6

Edited by author 30.09.2013 09:17

Edited by author 30.09.2013 09:17

Didn't notice that problem is not for Java, so I solved it. Maybe someone will need the solution. Solution without checking of stacks for emptiness.


import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
import java.util.StringTokenizer;

public class HelloWorld {

    public static void main (String[] args)     {
        HashMap<Integer,  ArrayList<Integer>> map = new HashMap<Integer, ArrayList<Integer>>();
        Scanner reader = new Scanner(System.in);
        int n = reader.nextInt();
        reader.nextLine();
        for (int i = 0; i < n; i++) {
            String command = reader.nextLine();
            StringTokenizer tokenizer = new StringTokenizer(command);
            command = tokenizer.nextToken();
            if ("PUSH".equals(command)) {
                int key = Integer.parseInt(tokenizer.nextToken());
                int value = Integer.parseInt(tokenizer.nextToken());
                if (map.containsKey(key)) {
                    map.get(key).add(0, value);
                }
                else {
                    map.put(key, new ArrayList<Integer>());
                    map.get(key).add(0, value);
                }
            }
            else if ("POP".equals(command)) {
                int key = Integer.parseInt(tokenizer.nextToken());
                if (map.containsKey(key)) {
                    System.out.println(map.get(key).get(0));
                    map.get(key).remove(0);
                }
            }

        }
    }
}

Edited by author 31.10.2016 23:24

Edited by author 31.10.2016 23:26

Here is my not accepted solution:
#include<iostream>
#include<string>
using namespace std;

double P(int power[],string names[],int a,int n,int video_power,int on_or_off[],int w,int h){
a=video_power;
for(int i=0;i<n;i++){if(on_or_off[i]==1){a/=power[i];}}
double res=a/(w*h);
return res;
}

int main()
{

int n;
cin>>n;
string * names = new string[n];
int * power = new int[n];
for (int i=0;i<n;i++){
cin>>names[i];
cin>>power[i];
}

int w,h,video_power;
cin>>w>>h>>video_power;

double results[100];
int on_or_off[100];
for(int i=0;i<n;i++){on_or_off[i]=1;}
int e=0;

int a=video_power;
for (int i=0; i<n;i++){
a/=power[i];
}
double prodyctivity=a/(w*h);
results[e]=prodyctivity;e++;

int m;
cin>>m;
for(int i=0; i<m; i++){
string change,change_name;
cin>>change;
if(change[0]!='R'){
cin>>change_name;
for(int y=0;y<n;y++){
    if(change=="On" && change_name==names[y]){
        on_or_off[y]=1;
        prodyctivity=P(power,names,a,n,video_power,on_or_off,w,h);
        results[e]=prodyctivity;e++;
        }
    else if(change=="Off" && change_name==names[y]){
        on_or_off[y]=0;
        prodyctivity=P(power,names,a,n,video_power,on_or_off,w,h);
        results[e]=prodyctivity;e++;
        }
}
}
else{
int a1,a2;
cin>>a1>>a2;
w=a1;h=a2;
prodyctivity=P(power,names,a,n,video_power,on_or_off,w,h);
results[e]=prodyctivity;e++;
}
}

for(int i=0;i<m+1;i++){
    if(results[i]<10){cout<<"Slideshow"<<endl;}
    else if(results[i]>=60){cout<<"Perfect"<<endl;}
    else{cout<<"So-so"<<endl;}
}
}

Please!

If the plane fly across a edge and it will produce a point then produce a new block so the answer add one.
If gcd(M, N) = 1, it means there is no situation that the plane fly across the crossroads.
See from left to right it produce M point and see from north to south it produce N point,
but one is calculate twice, so there are M + N - 1 points and Ans is M + N - 1.

I've perused over some of the other solutions here, tried to come up with something original, but I'm having a difficult time understanding why it won't stop letting me input in Eclipse, and how to deal with that...Here's my code:

import java.util.*;

public class ReverseRoot
{//start class
    public static void main(String[] args)
    {//start main
        Scanner in = new Scanner(System.in);
        ArrayList<Long> array = new ArrayList<Long>();
        array.add(in.nextLong());

        while(in.hasNextLong())
        {
            array.add(in.nextLong());
        }
        in.close();

        for (int i = array.size(); i > 0; i--)
            System.out.printf("%.4f%n", Math.sqrt((double)array.get(i)));
    }//end main
}//end class

s1 = ['1 2 3 4 5',
      '1 4 3 2 5',
      '1 3 2 4 5',
      '1 3 4 2 5']

think about :

F(N,M)  =  F(N-? , M- ? )

and give the initial values.

#include<stdio.h>
#include<string.h>
int f(int n,int m)
{
    if(n==1||m==1)
        return 0;
    if(n==2&&m>1)
        return 2;
    if(n==3&&m==2)
        return 3;
    if(m>2&&n>2)
    return f(n-2,m-2)+4;

}
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    printf("%d",f(n,m));
    fflush(stdin);
    getchar();
    return 0;
}

* The minimal height of the wall is set to 1 (previously 0).
* The total number of bricks is set to be between 1 and 100 000 (previously, not more than 1 000 000, but there were no such tests).
* New tests have been added.
* All accepted solution have been rejudged, 98 authors have lost their AC.

[code deleted]

Edited by moderator 19.11.2019 23:13

It looks as first input number (sequence length) isn't needed for solution.

int a[10]={-1,1,2,2,1,-1,-2,-2,-1,1};
...
for(i=1;i<=8;i++) if(x+a[i-1]>0 && x+a[i-1]<9 && y+a[i+1]>0 && y+a[i+1]<9) h++;

DO IT!!! :)