anagram
Conversion of a Regular Expression to NFA Algorithm Source code C programmingCS342 Compiler Lab Source code Algorithm C Programming #include#includevoid main(){char reg[20];int q[20][3],i,j,len,a,b;clrscr();for(a=0;a
#include #include int main() { int i,j,k,l,m,n,p,q,temp,fno,secno,thirdno; int arr[1][9]; cout<<"---------------------------------------ABRAR TARIQ-------------------------------"< cout<<"Here are the all possible combinations of the triad numbers between 1 to 1000 ! cheers !"< int a=0; for(i=100;i<=1000;i++) { for(j=i+1;j<=1000;j++) { if(j==2*i) { fno=i; secno=j; } if(j==3*i) { thirdno=3*i; } } k=fno/100; temp=fno%100; l=temp/10; m=temp%10; arr[0][0]=k; arr[0][1]=l; arr[0][2]=m; k=secno/100; temp=secno%100; l=temp/10; m=temp%10; arr[0][3]=k; arr[0][4]=l; arr[0][5]=m; k=thirdno/100; temp=thirdno%100; l=temp/10; m=temp%10; arr[0][6]=k; arr[0][7]=l; arr[0][8]=m; for(n=0;n<1;n++) { for(p=0;p<9;p++) { for(q=p+1;q<9;q++) { if(arr[n][p]==arr[n][q]arr[n][p]==0arr[n][q]==0) { a=1; } if(a==1) break; } if(a==1) { a=0; break; } } if(p==9&&q==9&&a==0) { cout<<<","<<<","<<< cout<<"---And---"< } } return 0; }
#include<iostream> #include<string.h> //#include<graphics.h> //#include<dos.h> #include<math.h> using namespace std; int main() { // clrscr(); int c=1,c1=4,c2=1,c3=1,r[2000][10],c4=1,c5=1,p[2000][10],q[2000][10],o[2000][10],t[2000][10],s[2000][10],w[2000][10],i,l=1,j,k,m,u[2000][10],a[2000][10],v,e=0,g=1,f=1,h=0,b[2000][10],z[2000][10],y[6000][5],x[2000][10]; cout<<"Enter 0 for not selecting and 1 for selecting\n"; for(i=0;i<=15;i++) { cout<<"Enter the value for "<<i<<":"; cin>>v; if(v==1) { a[c][5]=i; m=i; while(c1!=0) { a[c][c1]=m%2; m=m/2; c1--; } c++; c1=4; } } for(i=1;i<c;i++) { for(j=1;j<c;j++) { for(k=1;k<=4;k++) { if(a[i][k]+a[j][k]==1) e++; } if(e==1) { for(k=1;k<=4;k++) { if(a[i][k]+a[j][k]!=1) b[f][g]=a[i][k]; else b[f][g]=3; g++; } b[f][5]=a[i][5]; b[f][6]=a[j][5]; g=1; f++; h++; } e=0; } if(h==0) { for(k=1;k<=4;k++) x[c2][k]=a[i][k]; c2++; } h=0; } c=0; for(i=1;i<f;i++) { for(j=1;j<i;j++) { for(k=1;k<=4;k++) { if(b[i][k]==b[j][k]) c++; } if(c!=4) e++; c=0; } if(e==i-1) { for(k=1;k<=6;k++) u[h+1][k]=b[i][k]; h++; } e=0; } for(i=1;i<=h;i++) { if(u[i][7]!=4) { for(j=1;j<=h;j++) { if(u[j][7]!=4) { if((u[i][6]==u[j][5])(u[i][6]==u[j][5])(u[i][5]==u[j][5])(u[i][5]==u[j][6])) { for(k=1;k<=h;k++) { if(u[j][6]==u[k][5]u[j][6]==u[k][6]u[j][5]==u[k][5]u[j][5]==u[k][6]) { for(e=1;e<=h;e++) { if(i!=j&&i!=k&&i!=e&&j!=k&&j!=e&&k!=e) { if(u[k][6]!=u[e][5]&&u[k][6]!=u[e][6]&&u[k][6]!=u[e][6]&&u[k][6]!=u[e][5]) u[j][7]=4; } } } } } } } } } cout<<endl; f=h+1; e=0;h=0;g=1;l=1;c=1; for(i=1;i<f;i++) { for(j=1;j<f;j++) { for(k=1;k<=4;k++) { if(u[i][k]+u[j][k]==1) e++; if((u[i][k]+u[j][k]==3)(u[i][k]+u[j][k]==4)) e=2; } if(e==1) { for(k=1;k<=4;k++) { if(u[i][k]+u[j][k]!=1&&u[i][k]+u[j][k]!=6) z[l][g]=u[i][k]; else z[l][g]=3; g++; } g=1; l++; h++; } e=0; } if(h==0) { for(k=1;k<=7;k++) w[c3][k]=u[i][k]; c3++; } h=0; } c=0; for(i=1;i<l;i++) { for(j=1;j<i;j++) { for(k=1;k<=4;k++) { if(z[i][k]==z[j][k]) c++; } if(c!=4) e++; c=0; } if(e==i-1) { for(k=1;k<=4;k++) t[h+1][k]=z[i][k]; h++; } e=0; } l=h+1; e=0;h=0;g=1;m=1;c=1;c1=4; for(i=1;i<l;i++) { for(j=i+1;j<l;j++) { for(k=1;k<=4;k++) { if(t[i][k]+t[j][k]==1) e++; if(t[i][k]+t[j][k]==3t[i][k]+t[j][k]==4) e=2; } if(e==1) { for(k=1;k<=4;k++) { if(t[i][k]+t[j][k]!=1&&t[i][k]+t[j][k]!=6) y[m][g]=t[i][k]; else y[m][g]=3; g++; } g=1; m++; h++; } e=0; } if(h==0) { for(k=1;k<=4;k++) r[c4][k]=t[i][k]; c4++; } h=0; } c=0;h=0; for(i=1;i<m;i++) { for(j=1;j<i;j++) { for(k=1;k<=4;k++) { if(y[i][k]==y[j][k]) c++; } if(c!=4) e++; c=0; } if(e==i-1) { for(k=1;k<=4;k++) s[h+1][k]=y[i][k]; h++; } e=0; } f=h+1; e=0;h=0;g=1;l=1;c=1; for(i=1;i<f;i++) { for(j=1;j<f;j++) { for(k=1;k<=4;k++) { if(s[i][k]+s[j][k]==1) e++; if((s[i][k]+s[j][k]==3)(s[i][k]+s[j][k]==4)) e=2; } if(e==1) { for(k=1;k<=4;k++) { if(s[i][k]+s[j][k]!=1&&s[i][k]+s[j][k]!=6) q[l][g]=s[i][k]; else q[l][g]=3; g++; } g=1; l++; h++; } e=0; } if(h==0) { for(k=1;k<=4;k++) p[c5][k]=s[i][k]; c5++; } h=0; } c=0; for(i=1;i<l;i++) { for(j=1;j<i;j++) { for(k=1;k<=4;k++) { if(q[i][k]==q[j][k]) c++; } if(c!=4) e++; c=0; } if(e==i-1) { for(k=1;k<=4;k++) o[h+1][k]=q[i][k]; h++; } e=0; } for(i=1;i<=h;i++) { cout<<1; } cout<<endl; for(i=1;i<c2;i++) { if(x[i][1]==0) cout<<"A'"; if(x[i][1]==1) cout<<"A"; if(x[i][2]==0) cout<<"B'"; if(x[i][2]==1) cout<<"B"; if(x[i][3]==0) cout<<"C'"; if(x[i][3]==1) cout<<"C"; if(x[i][4]==0) cout<<"D'"; if(x[i][4]==1) cout<<"D"; cout<<"+"; } c=0; for(i=1;i<c3;i++) { if(w[i][7]!=4) { if(w[i][1]==0) cout<<"A'"; if(w[i][1]==1) cout<<"A"; if(w[i][2]==0) cout<<"B'"; if(w[i][2]==1) cout<<"B"; if(w[i][3]==0) cout<<"C'"; if(w[i][3]==1) cout<<"C"; if(w[i][4]==0) cout<<"D'"; if(w[i][4]==1) cout<<"D"; cout<<"+"; } } cout<<endl; c=0; for(i=1;i<c4;i++) { for(j=1;j<=h;j++) { if(((o[j][1]-r[i][1])==2(o[j][1]-r[i][1])==3(o[j][1]-r[i][1])==0)&&((o[j][2]-r[i][2])==2(o[j][2]-r[i][2])==3(o[j][2]-r[i][2])==0)&&((o[j][3]-r[i][3])==2(o[j][3]-r[i][3])==0(o[j][3]-r[i][3])==3)&&((o[j][4]-r[i][4])==2(o[j][4]-r[i][4])==0(o[j][4]-r[i][4])==3)) c++; } for(j=1;j<c5;j++) { if(((p[j][1]-r[i][1])==2(p[j][1]-r[i][1])==3(p[j][1]-r[i][1])==0)&&((p[j][2]-r[i][2])==2(p[j][2]-r[i][2])==3(p[j][2]-r[i][2])==0)&&((p[j][3]-r[i][3])==2(p[j][3]-r[i][3])==0(p[j][3]-r[i][3])==3)&&((p[j][4]-r[i][4])==2(p[j][4]-r[i][4])==0(p[j][4]-r[i][4])==3)) c++; } if(c==0) { if(r[i][1]==0) cout<<"A'"; if(r[i][1]==1) cout<<"A"; if(r[i][2]==0) cout<<"B'"; if(r[i][2]==1) cout<<"B"; if(r[i][3]==0) cout<<"C'"; if(r[i][3]==1) cout<<"C"; if(r[i][4]==0) cout<<"D'"; if(r[i][4]==1) cout<<"D"; cout<<"+"; } c=0; } cout<<endl; for(i=1;i<c5;i++) { if(p[i][1]==0) cout<<"A'"; if(p[i][1]==1) cout<<"A"; if(p[i][2]==0) cout<<"B'"; if(p[i][2]==1) cout<<"B"; if(p[i][3]==0) cout<<"C'"; if(p[i][3]==1) cout<<"C"; if(p[i][4]==0) cout<<"D'"; if(p[i][4]==1) cout<<"D"; cout<<"+"; } // getch (); }
K j equals j when K is 1 or j is 0.
#include<stdio.h> void main() { int a[5],b[5]; int c[10]; int i,j,temp,k; printf("\n enter the elements of array A:=\n"); for(i=0;i<5;i++) scanf("%d",&a[i]); printf("\n enter the elements of array B:=\n"); for(i=0;i<5;i++) scanf("%d",&b[i]); for(i=1;i<5;i++) { for(j=0;j<5-i;j++) { if(a[j]>a[j+1]) { temp=a[j]; a[j]=a[j+1]; a[j+1]=temp; } } for(j=0;j<5-i;j++) { if(b[j]>b[j+1]) { temp=b[j]; b[j]=b[j+1]; b[j+1]=temp; } } } printf("\n the elements of the array A:=\n"); for(i=0;i<5;i++) printf("%d\t",a[i]); printf("\n the elements of the array B:=\n"); for(i=0;i<5;i++) printf("%d\t",b[i]); i=0,j=0,k=0; while(i<5&&j<5) { if(a[i]>b[j]) c[k++]=b[j++]; if(a[i]<b[j]) c[k++]=a[i++]; } if(i<5&&j==5) while(i<5) c[k++]=a[i++]; if(i==5&&j<5) while(j<5) c[k++]=b[j++]; printf("\n the elements of the sorted merged array C:=\n"); for(i=0;i<10;i++) printf("%d\t",c[i]); }
The player with J, 10, since he has an Ace high straight. Also called "Broadway".
red Q, red Q, black Q, black Q, J, 6, 7, 3, K, 2, 5, 2, 5, J, 7, K, 6, 7, 3, K, 6, 7, 3, K, 2, 5, 10, 10, 10, 10, J, 4, 4, 8, 8, 8, 4, 9, 9, 9, 9, 8, A, A, A, A, "What The Hell", J, 2, 3, 4, 5, 6 (2-6 need to be spades)
Yes, that is a straight. A straight consists of five cards of consecutive rank. The Ace can be used as either the lowest or the highest card, (i.e., A, 2, 3, 4, 5 is a straight, as well as 10, J, Q, K, A)
The best Texas Hold'em hand is the royal flush, where you have a A, K, Q, J, and 10 of the same suit.
In poker, a straight is a hand that consists of five consecutive cards of any suit. The possible straights in poker are: A-2-3-4-5 (also known as a wheel or bicycle) 2-3-4-5-6 3-4-5-6-7 4-5-6-7-8 5-6-7-8-9 6-7-8-9-10 7-8-9-10-J 8-9-10-J-Q 9-10-J-Q-K 10-J-Q-K-A These are the ten possible straights in poker, with the highest being a 10-J-Q-K-A straight.
A red card means Hearts or Diamonds, a face card is a card which depicts a person, so J, Q, or K. Hence, the 6 red face cards are J, Q, K of hearts and J, Q, K of diamonds!
A, K, Q, J, 10 all of one suit - Royal Flush
No, a straight are five cards in sequence. Aces are high, deuces are low. They do not wrap around. Examples: A, K, Q, J, 10; A, 2, 3, 4, 5; 5, 6, 7, 8, 9.
J and Q .
Q = mcΔT Q = (1000 g)(4.18 J/(g*K))(5 K) Q = 20900 J Q = 20.9 kJ
Both players would have K, K, Q, J, 7. They would split the pot.
10, J, Q, K, A of the same suit without the use of a wild card(s). - Bdx