Find the power by O(lgn)

#include<bits/stdc++.h>
using namespace std;

#define out(x) cout << x << "\n"

// O(n) solution

//int power(int x, int y)
//{
//    if(y == 0) return 1;
//
//    if(y % 2 == 0)
//    {
//        int r = power(x, y/2);
//        return r*r;
//    }
//
//    else
//    {
//        return x * power(x, y/2) * power(x, y/2);
//    }
//}
//
//int main()
//{
//    int x,y;
//
//    while(cin >> x >> y)
//    {
//        cout << power(x, y) << "\n";
//    }
//}


// O(lgN) Solution

// when y is positive
// the function can be optimized to O(logn) by calculating power(x, y/2) only once and storing it.

//int power(int x, int y)
//{
//    int temp;
//
//    if(y == 0) return 1;
//
//    temp = power(x, y/2);
//
//    if(y % 2 == 0) return temp * temp;
//
//    else
//    {
//        return x * temp * temp;
//    }
//}
//
//int main()
//{
//    int x,y;
//
//    while(cin >> x >> y)
//    {
//        cout << power(x, y) << "\n";
//    }
//}


// O(lgn), when y is negative

float power(int x, int y)
{
    float temp=0.0;

    if(y == 0) return 1;

    temp = power(x, y/2);

    if(y % 2 == 0) return temp * temp;

    else
    {
        if(y > 0) return x * temp * temp;

        else
        {
            return temp * temp / x;
        }
    }
}

int main()
{
    int x,y;

    while(cin >> x >> y)
    {
        cout << power(x, y) << "\n";
    }
}

Longest Decreasing Subsequence

#include<bits/stdc++.h>
using namespace std;

const int inf = 9999;

int n, seq[100], ldsseq[100], I[100], L[100];

int BINARY_SEARCH(int lo, int hi, int itm)
{
    int mid;

    while(lo <= hi)
    {
        mid = (lo + hi) / 2;

        if(I[mid] < itm)
        {
            lo = mid + 1;
        }

        else
        {
            hi = mid - 1;
        }
    }

    return lo;
}

int LIS()
{
    int i, j , len=0 , r , Llen=0;

    for(i=0; i<n; i++)
    {
        r = BINARY_SEARCH(0, len, seq[i]);

        I[r] = seq[i];
        L[Llen++] = r;

        if(len < r)
        {
            len = r;
        }
    }

    //cout << "LDS Length: " << len;
    return len;
}

void find_squence( int mxlen)
{
    int i=0 , j;

    for(j=1; j<n; j++)
    {
        if(L[j] > L[i])
        {
            i = j;
        }
    }

    int top = L[i];

    top--;

    ldsseq[top] = seq[i];

    for(j=i-1; j>=0; j--)
    {
        if(seq[j] < seq[i] and L[j] == L[i]-1)
        {
            i=j;

            top--;

            ldsseq[top] = seq[i];
        }
    }

    cout << "Sequence: ";
    for(i=mxlen-1; i>=0; i--)
    {
        cout << ldsseq[i] << " ";
    }
}

int main()
{
    while(cin >> n)
    {
        int ilen=0;

        I[ilen++] = -inf;

        for(int i=0; i<n; i++)
        {
            cin >> seq[i];

            I[ilen++] = seq[i];
        }

        reverse(seq, seq+n);

        int rlen = LIS();

        find_squence(rlen);

        cout << "LDS Length = " << rlen << "\n";
    }

    return 0;
}

Longest Increasing Subsequence with O(N log K)

// LIS -- O(N log K)

#include<bits/stdc++.h>
#define sze 100

using namespace std;

const int inf=9999;

int seq[sze+5], lisseq[sze+5] , n , I[sze+5], L[sze+5];

int Binary_Search(int lo, int hi, int itm)
{
    int mid;

    while(lo <= hi)
    {
        mid = (lo + hi) / 2;

        if(I[mid] < itm)
        {
            lo = mid+1;
        }

        else
        {
            hi = mid-1;
        }
    }

    return lo;
}

int LIS()
{
    //memset(L, 1, sizeof(L));

    int i , r , len=0 , llen=0;

    for(i=0; i<n; i++)
    {
        r = Binary_Search(0, len, seq[i]);

        I[r] = seq[i];
        L[llen++] = r;

        if(len < r)
        {
            len = r;
        }
    }

    return len;
}

void find_sequence(int mxlen)
{
    int i=0 , j;

    for(j=1; j<n; j++)
    {
        if(L[i] < L[j])
        {
            i = j;
        }
    }

    int top = L[i];

    top--;

    lisseq[top] = seq[i];

    for(j=i-1; j>=0; j--)
    {
        if(seq[j] < seq[i] and L[j] == L[i] - 1)
        {
            i = j;

            top--;

            lisseq[top] = seq[i];
        }
    }

    cout << "LIS Sequence: ";

    for(i=0; i<mxlen; i++)
    {
        cout << lisseq[i] << " ";
    }
}

int main()
{
    while(cin >> n)
    {
        int ilen=0;

         I[ilen++] = -inf;

        for(int i=0;i<n;i++)
        {
            cin >> seq[i];

            I[ilen++] = seq[i];
        }

        int lislen = LIS();

        find_sequence(lislen);

        cout << "LIS Length: " << lislen << "\n";
    }

    return 0;
}

Find Total Number of Distinct Prime Factors

#include<bits/stdc++.h>

#define limit 100100

using namespace std;

typedef long long LL;
typedef bitset<limit>bs;

int distinctfactors[limit] = {0};

bs bit;

void distinct_Prime_factors()
{
    LL i,j;

    bit.set();
    bit[0] = bit[1] = 0;

    for(i=2; i<limit; i++)
    {
        if(bit[i])
        {
            for(j=i*2; j<limit; j+=i)
            {
                if(!(j % i))
                {
                    distinctfactors[j]+=1;
                }

                bit[j] = 0;
            }
        }
    }
    //for(i=2;i<10;i++)cout<<distinctfactors[i]<<" ";
}

int main()
{
    distinct_Prime_factors();
    cout << "Enter a number to find total distinct prime factor: ";
    int n;
    while(cin >> n)
    {
        cout << "Total Distinct Prime Factor = " << distinctfactors[n] << "\n";

        cout << "Enter a number to find total distinct prime factor: ";
    }

    return 0;
}

Find The Total Number of Prime Factors of a number

#include<bits/stdc++.h>

#define limit 100100

using namespace std;

typedef long long LL;
typedef bitset<limit>bs;

int totalfactors[limit];

bs bit;

void find_total_factors()
{
    LL i,j;
    bit.set();
    bit[0] = bit[1] = 0;

    for(i=2; i<limit; i++)
    {
        if(bit[i])
        {
            totalfactors[i] = 1;

            for(j=i*2; j<limit; j+=i)
            {
                totalfactors[j] = 1 + totalfactors[j / i];

                bit[j] = 0;
            }
        }
    }
}

int main()
{
    find_total_factors();
    cout<<"Enter number to find total factors of that number: ";
    int number;
    while(cin >> number)
    {
        cout<<"Total Factors of " << number << " is = " << totalfactors[number] << "\n";

        cout<<"Enter number to find total factors of that number: ";
    }

    return 0;
}

Big Integer Sum between Two Numbers

#include<bits/stdc++.h>

using namespace std;

void x_y(string a, string b)
{
    int alen=a.size(), blen=b.size(),i,j, sum, carry=0, num[1009], numlen=0;

    i=alen-1;
    j=blen-1;

    for(; i>=0; i--)
    {
        sum = a[i]-48;

        if(j>=0)
        {
            sum+=(b[j]-48);
            j--;
        }

        sum+=carry;

        if(sum > 10)
        {
            num[numlen]=sum%10;
            carry=sum/10;
        }
        else
        {
            num[numlen]=sum%10;
            carry=sum/10;
        }

        numlen++;
    }

//    if(carry)
//    {
//        cout << carry;
//    }

    if(carry)
    {
        num[numlen++]=carry;
    }

    for(i=numlen-1; i>=0; i--)
    {
        cout << num[i];
    }

    cout << endl;
}

void eql(string a, string b)
{
    int alen=a.size(), blen=b.size(), i, j, sum, carry=0, num[1009], numlen=0;

    i=alen-1;
    j=blen-1;

    for(; i>=0 and j>=0; i--,j--)
    {
        sum = (a[i]-48) + (b[j]-48);

        sum+=carry;

        if(sum > 10)
        {
            num[numlen]=sum%10;
            carry=sum/10;
        }

        else
        {
            num[numlen]=sum%10;
            carry=sum/10;
        }

        numlen++;
    }

    if(carry)
    {
        cout << carry;
    }

    for(i=numlen-1; i>=0; i--)
    {
        cout << num[i];
    }

    cout << endl;
}

int main()
{
    string x,y;

    while(cin >> x >> y)
    {
        if(y.size() > x.size())
        {
            swap(x,y);
        }

        if(x.size() > y.size())
        {
            x_y(x,y);
        }

        else
        {
            eql(x,y);
        }
    }

    return 0;
}

sscanf/sprintf (stdio.h/cstdio)

#include<bits/stdc++.h>

using namespace std;

#define pf printf

int main()
{
    //cout << "hello";

    char ch[20];
    gets(ch);

    int cid;
    char name[100];
    int solved;

    sscanf(ch, "%d %s %d", &cid,&name,&solved); // String theke data read korar jonno

    pf("%d %s %d\n",cid,name,solved);

    int n=1001;
    char cr[10];

    sprintf(cr, "%d", n); // kono kichuke Char e convert korte
    pf("%s",cr);

    return 0;
}