Practice Make Perfect!

Ukhuwah-created by unknown

dzikrina — Sun, 06 Feb 2011 06:21:39 +0000

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

dzikrina — Fri, 29 Oct 2010 10:31:34 +0000

//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

dzikrina — Sat, 23 Oct 2010 16:57:12 +0000

//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

dzikrina — Sat, 23 Oct 2010 14:46:30 +0000

//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++

dzikrina — Fri, 07 May 2010 10:03:21 +0000

#include
//#include

using namespace std;

typedef long long LL;

LL A,B,C;

LL recur(LL b){
//cout< 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 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<
}
//system("pause");
return 0;
}

practice contest 22/1/2010

dzikrina — Fri, 22 Jan 2010 11:18:32 +0000

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 int st = -1; for(int t=0; t int mindist, id; mindist = -1; ++ st, st %= n; while( ok[ st ] ) ++st, st %= n; for(int i=0; i 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 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 for (int i=0; i> sslot; cin >> seq; for (int k=(sslot-1); k garden [k]=1; } int ans = 0; for(int i=0; i 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 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

dzikrina — Fri, 22 Jan 2010 11:12:54 +0000

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

return 0;
}

tes

dzikrina — Mon, 21 Dec 2009 10:32:21 +0000

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

Item 1

Item 2

Item 3

Item 4

This is a big letter order list

Item 1

Item 2

Item 3

Item 4

This is a small letter order list

Item 1

Item 2

Item 3

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

dzikrina — Sat, 19 Dec 2009 07:38:42 +0000

// 1001 HELLO WORLD!

#include
using namespace std;

int main()
{
int num;

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

cout << C;
}
cout << endl;

system (“pause”);
return 0;
}

Quick Change

dzikrina — Sat, 19 Dec 2009 06:57:27 +0000

// QUICK CHANGE OR FAST CHANGE

#include
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 (BASE*QUARTER);
D = static_cast (BASE*DIME);
N = static_cast (BASE*NICKEL);
P = static_cast (BASE*PENNY);

cin >> numCase;

for (int i=0; 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;
}