Practice Make Perfect!

Ukhuwah-created by unknown

Posted by: dzikrina on: February 6, 2011


Bersama kita menjalin rasa ukhuwwah
Yang telah bersemi
Berawal dari rasa cinta Ilahi
Menjelma rasa cinta sejati

Seindah senyum manis
Matahari bersinar semarakkan pagi
Dedaunan melambai hijau nan berseri
Kicauan burung damaikan hati

Oo.. berpadulah,
hapuskan sgala rasa amarah
Pupus sudah rasa dendam
padamkan bara api permusuhan
Sesegar mawar bertaburan,
harum mewangi rasa persaudaraan…

Crazy Phytagoras

Posted by: dzikrina on: October 29, 2010

//checking phytagoras rectangle

#include
#include
#include

using namespace std;
int main()
{
int input[3];
int sum, C;

while (cin >> input[0] >> input[1] >> input[2])
{

if ( (input[0]== 0) && (input[1] == 0) && (input[2]== 0))
break;

sort( input, input+3 );

sum = ( (input[0]*input[0])+(input[1]*input[1]));
C = (input[2]*input[2]);

if ( sum == C)
cout << "right" << endl;
else
cout << "wrong" << endl;

}

return 0;

}

11388 gcd lcm

Posted by: dzikrina on: October 23, 2010

//http://uva.onlinejudge.org/external/113/11388.html
//http://github.com/andmej/acm/tree/master/11388%20-%20GCD%20LCM

//11388 PROBLEM d: gcd lcm
#include

using namespace std;

int main()
{
int t;
unsigned int g, l;

cin >> t;
while (t--)
{
cin >> g >> l;
if ( g > l || l % g != 0)
cout << "-1" << endl;
else
cout << g << " " << l << endl;

}

//system ("pause");
return 0;
}

11636

Posted by: dzikrina on: October 23, 2010

//http://uva.onlinejudge.org/external/116/11636.html
// 11636 Hello World!
#include
#include
using namespace std;

int main()

{
int mycase=0, N;
double k;

while (1)
{
cin >> N;
if (N <0)
break;
k = ceil(log(N)/log(2));
int i=0;
mycase ++;
cout <<"Case " << mycase << ": " << k << endl;
}
return 0;
}

BIG MODULUS C++

Posted by: dzikrina on: May 7, 2010

#include
//#include

using namespace std;

typedef long long LL;

LL A,B,C;

LL recur(LL b){
//cout<<b<<endl;
if(b==1) return A%C;
LL t1 = recur(b/2);
t1 = (t1*t1)%C;
if(b%2==1) t1 = (t1*A)%C;
return t1;
}

/*LL pow(LL X,int p){
LL ret = 1;
for(int i=0;i<p;++i) ret=(ret*X)%C;
return ret;
}

LL recur10(LL b){
if(b<10) return pow(A,b);
LL t1 = recur10(b/10);
t1 = pow(t1,10);
for(int i=0;i> T;

for (int i=0; i>A >> B >> C;
cout<<recur(B)<<endl;

}
//system("pause");
return 0;
}

practice contest 22/1/2010

Posted by: dzikrina on: January 22, 2010

http://acm.tju.edu.cn/toj/vcontest/contest5472.html

PROBLEM B
B. Serious Cow Tag

/*
TASK:
LANG: C++
USER: smilitude1
*/

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

struct data { int x,y; } ;

int main() {
data d[1005];
int n;
bool ok[1005]; SET( ok, 0 );
scanf("%d",&n);
for(int i=0; i<n; i++) scanf("%d %d",&d[i].x, &d[i].y);
int st = -1;

for(int t=0; t<n-1; t++) {
int mindist, id;
mindist = -1;
++ st, st %= n;
while( ok[ st ] ) ++st, st %= n;
for(int i=0; i<n; i++) if( !ok[i] && i != st ) {
int dist = ( d[i].x - d[st].x ) * ( d[i].x - d[st].x ) + ( d[i].y - d[st].y ) * ( d[i].y - d[st].y );
if( mindist == -1 || dist < mindist ) mindist = dist, id = i;
}
ok[ id ] = true;
}

for(int i=0; i<n; i++) if( !ok[i] ) printf("%d\n",i+1);

return 0;
}

PROBLEM C
The Flower Garden

#include
using namespace std;

int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);

int row,garden[10010],numr;
int sslot, seq, count=0;

cin >> row;
cin >> numr;

for (int j=0; j<row; j++) garden[j]=0;

for (int i=0; i> sslot;
cin >> seq;

for (int k=(sslot-1); k<row; k+=seq)
garden [k]=1;
}

int ans = 0;
for(int i=0; i<row; i++) if( garden[i] == 0 ) ans ++;
cout << ans << endl;

return 0;
}

PROBLEM D
Sudoku

#include
#include
#include
#include

#include
#include
#include
#include

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

int main() {
int total, nflower;

while( scanf("%d%d",&total, &nflower ) == 2 ) {
bool flag[total+1];
int a,b;
memset( flag, 0, sizeof( flag ));

for(int i=0; i<nflower; i++) {
scanf("%d%d",&a, &b );
for(int j=a; j<=total; j+=b) flag[j] = true;
}
int ans = 0;
for(int i=1; i<=total; i++) if(flag[i] == 0 ) ans++;
printf("%d\n",ans);
}

return 0;
}

PROBLEM E
RED AND BLACK

/*
TASK: TJU 1851
ALGO: backtracking
LANG: C++
USER: smilitude

TIME:
*/

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

#define REP(i,n) for(int i=0; i<(n); i++)
#define FOR(i,a,b) for( __typeof(b) i=(a); i=(b); i--)
#define SET(t,v) memset((t), (v), sizeof(t))
#define sz size()
#define pb push_back
#define i64 long long
#define ALL(x) x.begin(), x.end()

#define FORIT(i, m) for (__typeof((m).begin()) i=(m).begin(); i!=(m).end(); ++i)
#define REV(x) reverse( ALL( x ) )

#define IO freopen("","r",stdin); freopen("","w",stdout);
#define bug(x) cerr << __LINE__ <<" "<< #x " = " << x << endl

char g[10][10];
int used[3][3][10], row[9][10], col[9][10];
bool done;
int mid = 15;

void print() {
REP(i,9) {
REP(j,9) printf("%d", g[i][j] );
puts("");
}
// puts("");
}

void solve( int r, int c ) {
if( r == 9 ) {
done = 1;
print();
}
else if( c == 9 ) solve( r+1, 0 );
else if( g[r][c] != 0 ) solve( r, c+1 );
else {
int rr = r/3, cc = c/3;
FOR(i,1,9) {
if( row[r][i] || col[c][i] || used[rr][cc][i] ) continue;

row[r][i] = 1;
col[c][i] = 1;
used[rr][cc][i] = 1;

g[r][c] = i;
solve( r, c+1 );
if( done ) return;

row[r][i] = 0;
col[c][i] = 0;
used[rr][cc][i] = 0;
}
g[r][c] = 0;
}
}

int main() {

int t;

scanf("%d",&t);
while( t-- ) {
REP(i,9) scanf("%s",g[i]);

// building
SET( used, 0 ); SET( row, 0 ); SET( col, 0 );
REP(r,9) REP(c,9) {
if( g[r][c] == '0' ) g[r][c] = 0;
else {
g[r][c] -= '0';
row[r][ g[r][c] ] = 1;
col[c][ g[r][c] ] = 1;
used[ r/3 ][ c/3 ][ g[r][c] ] = 1;
}
}

// processing
done = 0;
solve(0,0);

}

return 0;
}


ALL THE CODES BY MY TUTORS, BRO IQRAM N BRO EMIR

IIUM PROGRAMMING TRAINING 2010

A. Simple Task

Posted by: dzikrina on: January 22, 2010

QUESTION

#include
#include
#include
#include

#include
#include
#include
#include

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

int a[205], b[205];
int na, nb;

int main() {
int t;
cin >> t;

while( t– ) {
cin >> na;
for(int i=0; i> a[i];
cin >> nb;
for(int i=0; i> b[i];

int ans = 0;
for(int i=0; i<na; i++)
for(int j=0; j<nb; j++)
if( (a[i] + b[j]) == 0 ) ans++;
cout << ans << endl;
}

return 0;
}

tes

Posted by: dzikrina on: December 21, 2009

fourth class

Learning 4

here are some sample of html expertise:
Here’s some big text

Here’s some smaller text

And here’s a new paragraph to round things out.

this text is center

this is the example of the paragraph and make it right

this is the example of the paragraph and make it left

this is the example of the paragraph and make it center

Satu dua tiga empat, ternyata tidak mau center oh oh
Satu dua tiga empat, yang ini mau center, yeah!

This is a default order list

  1. Item 1
  2. Item 2
  3. Item 3
  4. Item 4

This is a big letter order list

  1. Item 1
  2. Item 2
  3. Item 3
  4. Item 4

This is a small letter order list

  1. Item 1
  2. Item 2
  3. Item 3
  4. Item 4

This is a default unorder list

  • Item 1
  • Item 2

    • This is a nested unorder list

    • Item 3
    • Item 4

This is a square unorder list

  • Item 1
  • Item 2

    • This is a circle list

    • Item 3
    • Item 4


Problem 1001, TJU

Posted by: dzikrina on: December 19, 2009

// 1001 HELLO WORLD!

#include <iostream>
using namespace std;

int main()
{
int num;

while (scanf (“%d”, &num) != EOF)
{
char C = static_cast<char> (num);

cout << C;
}
cout << endl;

system (“pause”);
return 0;
}

Quick Change

Posted by: dzikrina on: December 19, 2009

// QUICK CHANGE  OR FAST CHANGE

#include <iostream>
using namespace std;

int main()
{
int numCase;
int change;
int quarter, dime, nickel, penny;
int Q, D, N, P;
int rem1, rem2, rem3;

#define QUARTER 0.25
#define DIME 0.10
#define NICKEL 0.05
#define PENNY 0.01
#define BASE 100

Q = static_cast <int> (BASE*QUARTER);
D = static_cast <int> (BASE*DIME);
N = static_cast <int> (BASE*NICKEL);
P = static_cast <int> (BASE*PENNY);

cin >> numCase;

for (int i=0; i<numCase; i++)
{
cin >> change;

quarter = (change/Q);
rem1 = (change%Q);

dime = (rem1/D);
rem2 = (rem1%D);

nickel = (rem2/N);
rem3 = (rem2%N);

penny = (rem3/P);

cout << i+1 <<” “<< quarter << ” QUARTER(S), ” << dime << ” DIME(S), ” << nickel <<” NICKEL(S), “
<< penny << ” PENNY(S)” << endl;
}
return 0;
}

Older Entries

 

February 2012
M T W T F S S
« Feb    
 12345
6789101112
13141516171819
20212223242526
272829  

Blog Stats

  • 607 hits

Top Clicks

  • None

Blog at WordPress.com.

Theme: Albeo by Design Disease.

Follow

Get every new post delivered to your Inbox.