SQUARE ROOT OF A NUMBER

SQUARE ROOT OF A NUMBER

#include
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
#include
#include
#include
using namespace std;

// This class represents a directed graph using adjacency list representation
class Graph
{
    int V;    // No. of vertices
    list *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[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 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::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
    {
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::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
{
scanf("%d %d",&a,&b);
g.addEdge(a-1,b-1);
}
if(g.BFS(0))
{
if(g.isCyclic())
         cout << "NO"<
     else
         cout << "YES"<
}
else
printf("NO\n");
return 0;
}

HOW MANY GAMES (SPOJ)

PROBLEM CODE :GAMES
12448. HOW MANY GAMES
Write the average score x as a reduced fraction x=pq. 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.

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
{
if(s[j]!='.')
num=10*num+(s[j]-'0');
}
pow=1;
if(flag==0){
for(i=0;i
{
pow=pow*10;
}
}
//printf("%d %d",num,pow);
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
new_length=len_txt+count*(len_pattern-len_replace)
else there is no need to resize the string .
*/


#include//stdio.h
#include//string.h
#include//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
        {
            if(q<=size&&a[q]==i)
            {
                k=0;
                while(k
                {
                    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;
}

DRUNKEN KNIGHT

In a bizarre game of chess ,knight was so drunk, that instead of his usual move he started walking straight. In every move Knight jumps on 2n steps forward (n is number of block that he had travelled so far from starting) but after that he has to take either 1 step forward or backward.
Now the Knight needs to get to position X so King (i.e. You) needs to decide the order of his backward or forward step in such a way that he can reach its destination in minimum number of steps. Remember he always travels in a straight line and the length of the board is infinite.
Input

The first line of the input contains an integer T denoting the number of test cases, for each test case enter value X ( i.e. destination)
Note : initially knight is at n = 1.
Output

For each test case the output should be string of numbers 1 & 2 where 1 denotes backward step and 2 denote the forward step
Note : for no solution print 0.
Constraints

1 ≤ T ≤ 100
1 ≤ X ≤ 10^10
Example

Input:
2
17
10
Output:
2111
0
Explanation

Case 1 : starting from n = 1 , knight moves to n = 3 ('2') , 5 ('1') , 9 ('1') , 17 ('1') i.e. string printed is 2 1 1 1
Case 2 : no solution is possible


SOLUTION--

#include
int main()
{
int t,i,len;
long long int num;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&num);
char str[40];
i=0;
//num=num-1;
if(num%2==0)
            printf("0\n");
        else{
while(num>0)
{
if(num%2==0)
str[i]='1';
else
str[i]='2';
i++;
num=num/2;
}
len=i-1;
for(i=len;i>0;i--)
printf("%c",str[i]);
printf("\n");
        }
}
return 0;
}

Euler Totient Function

SPOJ Problem Set (classical)

4141. Euler Totient Function

Problem code: ETF

English Vietnamese
In number theory, the totient  of a positive integer n is defined to be the number of positive integers less than or equal to n that are coprime to n.

Given an integer n (1 <= n <= 10^6). Compute the value of the totient .

Input

First line contains an integer T, the number of test cases. (T <= 20000)

T following lines, each contains an integer n.

Output

T lines, one for the result of each test case.

Example

Input:
5
1
2
3
4
5

Output:
1
1
2
2
4



EXPLAINATION--
http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=primeNumbers

(TOPCODER FORUM)

SOLUTION--

#include
int phi(int n)
{
     int result = n;
     int i;
       for(i=2;i*i <= n;i++) 
       { 
         if (n % i == 0) 
         result -= result / i; 
         while (n % i == 0) 
         n /= i; 
       } 
       if (n > 1)
       result -= result / n; 
       return result; 
}
int main()
{
    int t,num;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&num);
        printf("%d\n",phi(num));
    }
    return 0;
}

GCD

SPOJ Problem Set (classical)

2906. GCD2

Problem code: GCD2

Frank explained its friend Felman the algorithm of Euclides to calculate the GCD of two numbers. Then Felman implements it algorithm

int gcd(int a, int b)
{
if (b==0)
return a;
else
return gcd(b,a%b);
}
and it proposes to Frank that makes it but with a little integer and another integer that has up to 250 digits.
Your task is to help Frank programming an efficient code for the challenge of Felman.

Input

The first line of the input file contains a number representing the number of lines to follow. Each line consists of two number A and B (0 <= A <= 40000 and A <= B < 10^250).

Output

Print for each pair (A,B) in the input one integer representing the GCD of A and B.

Example

Input:
2
2 6
10 11


Output:
2
1

EXPLAINATION--
In mathematics, the greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder. For example, the GCD of 8 and 12 is 4.


SOLUTION---

#include
#include
int mod(char str[],int d)
{
    int r=0,i;
    for(i=0;str[i];i++)
    {
        r=10*r+(str[i]-'0');
        r=r%d;
    }
    return r;
}
int gcd(int a,int b)
{
    if(b==0)
        return a;
    else
        gcd(b,a%b);
}
int main()
{
    int t,a;
    char str[255];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&a);
        scanf("%s",str);
        if(a==0)
            printf("%s\n",str);
        else if(str[0]=='0')
            printf("%d\n",a);
        else
        {
            printf("%d\n",gcd(a,mod(str,a)));   
        }
    }
    return 0;
}