Hexagonal Board SPOJ

7191 Hexagonal Board  
Problem Code HEXBOARD

#include<stdio.h>
int main()
{
int n,a=3,d=4,val,i,j;
scanf("%d",&n);
while(n!=-1)
{
val=a+(n-1)*d;
val=val-1;
int s[val+1][val+1];
for(i=0;i<=val;i++)
{
for(j=0;j<=val;j++)
s[i][j]=0;
}
s[0][val/2]=95;
for(i=1;i<=val/2;i++)
{
for(j=0;j<=val;j++)
{
if(s[i-1][j]==95)
{
s[i][j-1]=47;
s[i][j+1]=92;
}
if(s[i-1][j]==47){
//if(j-1>=0)
//s[i-1][j-1]=95;
//else
s[i][j]=92;
}
if(s[i-1][j]==92){
//if(j+1<=val)
//s[i-1][j+1]=95;
//else
s[i][j]=47;
}
}
//second loop//
for(j=0;j<=val;j++)
{
if(s[i][j]==47)
{
if(j-1>=0)
s[i][j-1]=95;
}
if(j+2<=val)
{
if(s[i][j]==92&&s[i][j+2]==47)
s[i][j+1]=95;
}
if(s[i][j]==92)
{
if(j+1<=val)
s[i][j+1]=95;
}
}
}
/*for(i=val/2+1;i<=val;i++)
{
for(j=0;j<=val;j++)
{
if(s[i][j]!=0)
printf("%c",s[i][j]);
else
printf(" ");
}
printf("\n");
}*/
for(i=0;i<=val/2;i++)
{
for(j=0;j<=val;j++)
{
if(s[i][j]!=0)
{
if(s[i][j]==47)
s[val-i+1][j]=92;
//printf("%c",92);
else if(s[i][j]==92)
s[val-i+1][j]=47;
//printf("%c",47);
else if(s[i][j]==95)
s[val-i][j]=95;
//printf("%c",s[i][j]);
}
// else
//printf(" ");
}
//printf("\n");
}
for(i=0;i<=val;i++)
{
for(j=0;j<=val;j++)
{
if(s[i][j]!=0)
printf("%c",s[i][j]);
else
printf(" ");
}
printf("\n");
}
printf("***\n");
scanf("%d",&n);
}
return 0;
}

SQUARE ROOT OF A NUMBER

#include<stdio.h>
float squareRoot(float n)
{
/*We are using n itself as initial approximation
This can definitely be improved */
float x = n;
float y = 1;
float e = 0.000001; /* e decides the accuracy level*/
while(x - y > e)
{
x = (x + y)/2;
y = n/x;
}
return x;
}

/* Driver program to test above function*/
int main()
{
int n;
scanf("%d",&n);
printf ("Square root of %d is %f", n, squareRoot(n));
getchar();
}

SPOJ::IS IT TREE

Q.) given an unweighted and undirected graph write a program to check is it a tree or not.
input:
In first line contain n (numbers of nodes) and m (number of edges)
then m lines contain m egdes of grapg (two integer u and v)
1<=N<=10000 ,1<=M<=20000
output:
print YES if it is a tree else print NO.

Just we have to check the graph is connected and have no cycle then it will be tree else no.
----------------------------------------------------------------

#include<iostream>
#include<stdio.h>
#include <list>
#include<limits.h>
using namespace std;

// This class represents a directed graph using adjacency list representation
class Graph
{
    int V;    // No. of vertices
    list<int> *adj;    // Pointer to an array containing adjacency lists
    bool isCyclicUtil(int v, bool visited[], bool *rs);
public:
    Graph(int V);  // Constructor
    void addEdge(int v, int w); // function to add an edge to graph
    bool BFS(int s);  // prints BFS traversal from a given source s
    bool isCyclic();    // returns true if there is a cycle in this graph
};

Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}

void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w); // Add w to v’s list.
}

bool Graph::BFS(int s)
{
    // Mark all the vertices as not visited
    bool *visited = new bool[V];
    for(int i = 0; i < V; i++)
        visited[i] = false;

    // Create a queue for BFS
    list<int> queue;

    // Mark the current node as visited and enqueue it
    visited[s] = true;
    queue.push_back(s);

    // 'i' will be used to get all adjacent vertices of a vertex
    list<int>::iterator i;

    while(!queue.empty())
    {
        // Dequeue a vertex from queue and print it
        s = queue.front();
       // cout << s << " ";
        queue.pop_front();

        // Get all adjacent vertices of the dequeued vertex s
        // If a adjacent has not been visited, then mark it visited
        // and enqueue it
        for(i = adj[s].begin(); i != adj[s].end(); ++i)
        {
            if(!visited[*i])
            {
                visited[*i] = true;
                queue.push_back(*i);
            }
        }
    }
    for(int k=0;k<V;k++)
    {
if(visited[k]==false)
return false;
    }
    return true;
}
bool Graph::isCyclicUtil(int v, bool visit[], bool *recStack)
{
    if(visit[v] == false)
    {
        // Mark the current node as visited and part of recursion stack
        visit[v] = true;
        recStack[v] = true;

        // Recur for all the vertices adjacent to this vertex
        list<int>::iterator i;
        for(i = adj[v].begin(); i != adj[v].end(); ++i)
        {
            if ( !visit[*i] && isCyclicUtil(*i, visit, recStack) )
                return true;
            else if (recStack[*i])
                return true;
        }

    }
    recStack[v] = false;  // remove the vertex from recursion stack
    return false;
}

// Returns true if the graph contains a cycle, else false.

bool Graph::isCyclic()
{
    // Mark all the vertices as not visited and not part of recursion
    // stack
    bool *visit = new bool[V];
    bool *recStack = new bool[V];
    for(int i = 0; i < V; i++)
    {
        visit[i] = false;
        recStack[i] = false;
    }

    // Call the recursive helper function to detect cycle in different
    // DFS trees
    for(int i = 0; i < V; i++)
        if (isCyclicUtil(i, visit, recStack))
            return true;

    return false;
}
int main()
{
int j,n,m,a,b;
scanf("%d %d",&n,&m);
Graph g(n);
for(j=0;j<m;j++)
{
scanf("%d %d",&a,&b);
g.addEdge(a-1,b-1);
}
if(g.BFS(0))
{
if(g.isCyclic())
        cout << "NO"<<endl;
    else
        cout << "YES"<<endl;
}
else
printf("NO\n");
return 0;
}

DIARY

QUESTION- http://www.spoj.com/problems/GCPC11F

SOURCE CODE-


//http://www.spoj.com/problems/GCPC11F/
#include<stdio.h>
#include<string.h>
int main()
{
    int t,i,count,shift,max;
    char s[2000];
    scanf("%d",&t);
        getchar();
    while(t--)
    {
        gets(s);
        char p[]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
        int hash[26];
        for(i=0;i<26;i++)
                hash[i]=0;
        int len=strlen(s);
        for(i=0;i<len;i++)
        {
            if(s[i]!=' ')
            {
                hash[s[i]-65]++;
            }
        }
        max=0;
        count=0;
        int pos;
        for(i=0;i<26;i++){
                if(hash[i]>max){
                        max=hash[i];
                        pos=i;
                        }
                //printf("%c %d ",i+65,hash[i]);
        }
        for(i=0;i<26;i++)
        {
                if(hash[i]==max)
                        count++;
        }
        int d;
        if(count>1)
                printf("NOT POSSIBLE\n");
        else
        {
                //printf("%d ",pos);
                shift=pos-4;
                if(shift<0)
                        shift=shift+26;
                printf("%d ",shift);
                for(i=0;i<len;i++)
                {
                        if(s[i]!=' ')
                        {
                                d=s[i]-65;
                                d=d-shift;
                                //printf("%d ",d);
                                if(d<0)
                                        d=d+26;
                                s[i]=p[d];
                        }
                        
                }
                printf("%s\n",s);
        }
    }
    return 0;
}

HOW MANY GAMES (SPOJ)

EXPLAINATION//


Write the average score x as a reduced fraction x=p/q. This means that p and q are integers, that q is positive and that q is minimal (or, equivalently, that p and q have no nontrivial common factor). Then the player can have played any multiple of q games hence the minimum number of games the player should have played is q.

When x=−30.25, note that −30.25=−1214 and −121 and 4 have no common factors except +1 and −1, hence the minimum number of games is indeed 4.

basically we have to convert the avg into fraction part p/q then calculate the gcd(p,q) 
answer will be q/gcd(p,q)



SOLUTION//


#include<stdio.h>
int gcd(int a,int b)
{
    if(b==0)
        return a;
    else
        gcd(b,a%b);
}
int main()
{
int t,num,count,i,j,len,pow,flag;
char s[50];
scanf("%d",&t);
while(t--)
{
scanf("%s",s);
len=strlen(s);
count=0,flag=1;
for(j=len-1;j>=0;j--)
{
if(s[j]=='.'){
flag=0;
break;
}
else
count++;
}
num=0;
for(j=0;j<len;j++)
{
if(s[j]!='.')
num=10*num+(s[j]-'0');
}
pow=1;
if(flag==0){
for(i=0;i<count;i++)
{
pow=pow*10;
}
}
printf("%d\n",pow/gcd(num,pow));
}
return 0;
}

MICROSOFT INTERVIEW QUESTION

/* QUESTION--
//http://www.careercup.com/question?id=16392679
The problem is given a string s1= "ABCDBCCDABCD". and a pattern "BC". we have to replace this pattern with other string ("UVW" or "U"or "uv"). Do this without creating new string.
Take the case to replace "BC" with following
a) "uvw" s1=AUVWDUVWCDAUVWD .
b) "U" s1=AUDUCDAUD .
c) "UV" s1=AUVDUVCDAUVD .

*/
/*Explaination
USING THE KMP algo the times of occurence and position of given pattern in the text then stored the position of the pattern  in two arrays (array a and array b ) 
for e.g text="Chandan"
pat="an";
replace="QQQ";
then a[0]=2,a[1]=5 and b[0]=3,b[1]=6
array containing the position of starting index postion of pattern and array b contain the ending position of the pattern .
now if using the KMP algo count the number of occurence of pattern in text.
and resize the string 
if(len_pattern<len_replace)
new_length=len_txt+count*(len_pattern-len_replace)
else there is no need to resize the string .
*/


#include<stdio.h>//stdio.h
#include<string.h>//string.h
#include<stdlib.h>//stdlib.h
int a[400];//ARRAYS TO STORE THE POSITION OF THE PATTERN IN THE STRING
int b[400];
int q=0,p=0;
void computeLPSArray(char *pat, int M, int *lps);

int KMPSearch(char *pat, char *txt)
{
int count=0;
    int M = strlen(pat);
    int N = strlen(txt);

    // create lps[] that will hold the longest prefix suffix values for pattern
    int *lps = (int *)malloc(sizeof(int)*M);
    int j  = 0;  // index for pat[]

    // Preprocess the pattern (calculate lps[] array)
    computeLPSArray(pat, M, lps);

    int i = 0;  // index for txt[]
    while(i < N)
    {
      if(pat[j] == txt[i])
      {
        j++;
        i++;
      }

      if (j == M)
      {
count++;
        //printf("%d\n", i-j);
        a[q]=i-j;
        b[p]=(i-j+M-1);
        //printf("%d %d\n",a[q],b[p]);
        q++,p++;

        j = lps[j-1];
      }

      // mismatch after j matches
      else if(pat[j] != txt[i])
      {
        // Do not match lps[0..lps[j-1]] characters,
        // they will match anyway
        if(j != 0)
         j = lps[j-1];
        else
         i = i+1;
      }
    }
return count;
    //free(lps); // to avoid memory leak
}

void computeLPSArray(char *pat, int M, int *lps)
{
    int len = 0;  // lenght of the previous longest prefix suffix
    int i;

    lps[0] = 0; // lps[0] is always 0
    i = 1;

    // the loop calculates lps[i] for i = 1 to M-1
    while(i < M)
    {
       if(pat[i] == pat[len])
       {
         len++;
         lps[i] = len;
         i++;
       }
       else // (pat[i] != pat[len])
       {
         if( len != 0 )
         {
      // This is tricky. Consider the example AAACAAAA and i =7.
           len = lps[len-1];

           // Also, note that we do not increment i here
         }
         else // if (len == 0)
         {
           lps[i] = 0;
           i++;
         }
       }
    }
}
int main()
{
    char txt[500],pat[500],replace[100];
    scanf("%s",txt);
    scanf("%s",pat);
    scanf("%s",replace);
    int len_txt=strlen(txt);
    int len_pat=strlen(pat);
    int len_replace=strlen(replace);
    int count=KMPSearch(pat, txt);

    int i,j,k,new_length;
    if(len_replace>=len_pat)
    {
        int increase_size=(len_replace*count)-(count*len_pat);
         new_length=len_txt+increase_size;
        txt[new_length];
        
        p=p-1;
       
        j=new_length-1;
        for(i=len_txt-1;i>=0;i--)
        {
            if(p>=0&&b[p]==i)
            {
                    k=len_replace-1;
                    while(k>=0)
                    {
                        txt[j]=replace[k];
                        k--;
                        j--;
                    }
                    p--;
                    i=i-len_pat+1;
            }
            else
            {
                txt[j]=txt[i];
                j--;
            }
        }
        printf("%s",txt);
    }
    else
    {
        int size=q-1;
        int decrease_size=(count*len_pat)-(count*len_replace);
        new_length=len_txt-decrease_size;
        j=0;
q=0;
        for(i=0;i<len_txt;i++)
        {
            if(q<=size&&a[q]==i)
            {
                k=0;
                while(k<len_replace)
                {
                    txt[j]=replace[k];
                    k++;
                    j++;
                }
                q++;
                i=i+len_pat-1;
            }
            else
            {
                txt[j]=txt[i];
                j++;
            }
        }
txt[j]='\0';
        printf("%s",txt);
    }
    return 0;
}

Knuth (Fisher-Yates) algorithm

/* microsoft
Write a function that gets a number n and prints out a random list
of numbers 1..n to the screen. For example:

randlist(5) : 1 5 3 2 4
randlist(6) : 4 6 1 5 3 2

This should be truly random (uniformly spread) and with a O(n) complexity.
Every number should appear only once. Random(n) is given as a tool you can use to generate a
single random number between 1-n */
//algorithm knuth suffle algo//
#include<stdio.h>
#include<stdlib.h>
int main()
{
int n,i=0,temp=0;
scanf("%d",&n);
int ar[n];
for(i=0;i<n;i++)
ar[i]=i+1;
while(i>0)
{
int r = rand()%i;//which is similar to ->rand()%i;
        temp = ar[i-1];
        ar[i-1] = ar[r];
        ar[r] = temp;
        i-=1;
}
for(i=0;i<n;i++)
printf("%d ",ar[i]);
return 0;
}