SQLite Tools

There is given a link that provides a Tutorial for the SQLite database tools.
Link: SQLite Tutorial

Prepared by:
---------------
Pranta Sarker
Lecturer
Department of CSE
North East University Bangladesh
Alternate Email: prantacse04@gmail.com

wubi Ubuntu Installer

Ubuntu Desktop Installer by wubi

Allen Turing The Great Mathemacian

Life of Sir Allen Turing

Program to find last digit of n’th Fibonnaci Number

/*

It takes a while before it is noticeable. In fact, the series is just 60 numbers long and then it repeats the same sequence again and again all the way through the Fibonacci series – for ever. The series of final digits repeats with a cycle length of 60 (Refer this for explanations of this result).
So, instead of calculating the Fibonacci number again and again, pre-compute the units digit of first 60 Fibonacci number and store it in an array and use that array values for further calculations.

Link: Program to find last digit of n’th Fibonnaci Number

*/


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

#define sf scanf
#define pf printf

LL f[65] = {0};

LL fib(LL n)
{
    f[0] = 0;
    f[1] = 1;

    for (LL i = 2; i <= n; i++)
    {
        f[i] = (f[i-1] + f[i-2])%10;
    }

    return f[n];
}

LL findLastDigit(LL n)
{
    fib(60);

    return f[n%60];
}

int main()
{
    int t , tc=0;
    LL n;
    sf("%d", &t);
    while(t--)
    {
        memset(f , 0 , sizeof f);

        sf("%lld", &n);

        pf("Case %d: %lld is the last digit.\n", ++tc, findLastDigit(n));
    }

    return 0;
}

ACM ICPC Preliminary 2016 Team Practice Contest A.Multiplication Queries

Problem Link: A - Multiplication Queries 

//Next Codeforces Round #354 (Div. 2)
#include<bits/stdc++.h>

//#include<cstdio>
//#include<iostream>
//#include<algorithm>
//#include<vector>
//#include<cstring>
//#include<cmath>
//#include<map>

using namespace std;

#define fast ios_base::sync_with_stdio(false)
#define bfast cin.tie(0)
#define outs(x) cout << x << " "
#define outn(x) cout << x << "\n"
#define sf scanf
#define pf printf
#define nl puts("")
//#define i64 long long
#define high 50005

typedef long long LL;
typedef vector<int>vii;
typedef vector<LL>vll;

LL ar[high], br[high];
LL tree[high * 3] , m;

LL multiple(LL a, LL b , LL c)
{
    if(!b) return 0;

    LL x = multiple(a , b/2, c)%c;

    if(!(b&1))
    {
        return ((x%c) + (x%c))%c;
    }

    else
    {
        return ((a%c) + (x%c)+(x%c))%c;
    }
}

LL bigmod(LL a, LL b , LL c)
{
    if(!b) return 1;

    LL x = bigmod(a , b/2, c)%c;

    if(!(b&1))
    {
        return multiple(x , x , c);
    }

    else
    {
        return (multiple(a%c, multiple(x , x , c)%c , c))%c;
    }
}

void build(LL node, LL b , LL e)
{
    if(b == e)
    {
        tree[node] = br[b];
        return;
    }

    LL Left = node * 2;
    LL Right = (node * 2)+1;
    LL mid = (b + e) / 2;

    build(Left , b , mid);
    build(Right, mid+1, e);

    tree[node] = (tree[Left]%m * tree[Right]%m)%m;
}

LL query(LL node, LL b , LL e, LL i, LL j)
{
    if(i > e or j < b) return 1;

    if(b >= i and e <= j)
    {
        //cout << tree[node] << "; ";
        return tree[node];
    }

    LL Left = node * 2;
    LL Right = (node * 2) + 1;
    LL mid = (b + e) / 2;

    LL pek = query(Left, b , mid, i , j);
    LL pdui = query(Right, mid+1, e , i , j);

    //cout << pek << " " << pdui << "; ";

    return (pek%m * pdui%m)%m;
}

int main()
{
    fast;
    int t , tc=0;
    LL n , q , k , i , l , r, tmp=1 , p;
    //cin >> t;
    sf("%d", &t);
    while(t--)
    {
        //cin >> n >> q >> k >> m;
        //ans.clear();
        sf("%lld %lld %lld %lld", &n, &q, &k ,&m);

        for(i=1; i<=n; i++)
        {
            //cin >> ar[i];
            sf("%lld", &ar[i]);
        }

        br[0] = 1;

        for(i=1; i<=n; i++)
        {
            br[i] = bigmod(ar[i] , k , m);

            //br[i]= (br[i]%m * br[i-1]%m)%m;
        }

        build(1 , 1 , n);

        //for(i=1; i<=n; i++) cerr << tree[i] << " ";

        pf("Case %d:\n", ++tc);

        while(q--)
        {
            sf("%lld %lld", &l, &r);

            LL ans = query(1, 1, n, l , r);
            pf("%lld\n", ans%m);
        }

      }

    return 0;
}
 

LightOJ ACM ICPC Preliminary 2016 Team Practice Contest G. Chocolates

Problem Link: G.Chocolates


//Next Codeforces Round #354 (Div. 2)
#include<bits/stdc++.h>

//#include<cstdio>
//#include<iostream>
//#include<algorithm>
//#include<vector>
//#include<cstring>
//#include<cmath>
//#include<map>

using namespace std;

#define fast ios_base::sync_with_stdio(false)
#define bfast cin.tie(0)
#define outs(x) cout << x << " "
#define outn(x) cout << x << "\n"
#define sf scanf
#define pf printf
#define nl puts("")
//#define i64 long long

typedef long long LL;
typedef vector<int>vii;
typedef vector<LL>vll;

vll v;

int main()
{
    int t;
    LL k , m , n , i , sum=0 , x , ans=0 , tmp;
    sf("%d", &t);
    while(t--)
    {
        v.clear();

        sf("%lld %lld %lld", &k, &m , &n);

        for(i=0; i<k; i++)
        {
            sf("%lld", &x);
            v.push_back(x);
        }

        sort(v.rbegin(), v.rend()); //for(i=0; i<k; i++) cerr << v[i] << " ";

        sum = ans = 1;

        for(i=1; i<k; i++)
        {
            if(v[i-1] == v[i])
            {
                ans+=sum;
            }

            else
            {
                sum+=(m * (v[i-1] - v[i]));
                ans+=sum;
            }
        }

        //cerr << ans << "\n";

        if(ans > n)
        {
            pf("impossible\n");
        }

        else
        {
            tmp = (n - ans) / k;

            pf("%lld\n", (tmp * k) + ans);
        }
    }

    return 0;
}

Extended Euclid

#include<bits/stdc++.h>

using namespace std;

typedef pair<int, int>pii;

pii EGCD(int a, int b)
{
    if(!b)
    {
        return pii(1, 0);
    }

    else
    {
        pii d = EGCD(b, a%b);

        return pii(d.second, d.first - (a / b) * d.second);
    }
}

int main()
{
    int a, b;

    while(cin >> a >> b)
    {
        pii ans = EGCD(a,b);

        cout << ans.first << " " << ans.second << "\n";
    }

    return 0;
}