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Who Wants To Know What 52 C in a P W J?
52 Cards in a Pack Without Jokers
52 c in a p(w J)?
52 cards in a pack (without Jokers)
What does 52 c in a p w j stand for?
Answer to 52 wk on a pIt stands for 52 White Keys on a Piano!
What does 52 c in a p?
52 cards in a pack
52 c in a p no j?
hey tis is sandy, it means 52 cuts in piano
What are the letters for Bella's lullaby on virtual piano?
J j h f d h a p a s o p J j h f d a p
When did J. C. P. Miller die?
J. C. P. Miller died in 1981.
When was J. C. P. Miller born?
J. C. P. Miller was born in 1906.
2 equals P C J P 1708 equals C W F S P C?
2 popes called john paul 1708 christopher wren finishes st pauls cathedral
Which letters of the alphabet have a line of turning symmetry for less than 360?
B C D G J P R S U a b c d e f g j m n p q r s t and u * * * * * What? G, J, P, R, a, b, d, e, f, g, j ,m, n, p, q, r, t and u have no line of symmetry!
C program to calculate First and Follow?
#include<stdio.h> #include<conio.h> #include<string.h> void main() { char t[5],nt[10],p[5][5],first[5][5],temp; int i,j,not,nont,k=0,f=0; clrscr(); printf("\nEnter the no. of Non-terminals in the grammer:"); scanf("%d",&nont); printf("\nEnter the Non-terminals in the grammer:\n"); for(i=0;i<nont;i++) { scanf("\n%c",&nt[i]); } printf("\nEnter the no. of Terminals in the grammer: ( Enter e for absiline ) "); scanf("%d",&not); printf("\nEnter the Terminals in the grammer:\n"); for(i=0;i<nott[i]=='$';i++) { scanf("\n%c",&t[i]); } for(i=0;i<nont;i++) { p[i][0]=nt[i]; first[i][0]=nt[i]; } printf("\nEnter the productions :\n"); for(i=0;i<nont;i++) { scanf("%c",&temp); printf("\nEnter the production for %c ( End the production with '$' sign ) :",p[i][0]); for(j=0;p[i][j]!='$';) { j+=1; scanf("%c",&p[i][j]); } } for(i=0;i<nont;i++) { printf("\nThe production for %c -> ",p[i][0]); for(j=1;p[i][j]!='$';j++) { printf("%c",p[i][j]); } } for(i=0;i<nont;i++) { f=0; for(j=1;p[i][j]!='$';j++) { for(k=0;k<not;k++) { if(f==1) break; if(p[i][j]==t[k]) { first[i][j]=t[k]; first[i][j+1]='$'; f=1; break; } else if(p[i][j]==nt[k]) { first[i][j]=first[k][j]; if(first[i][j]=='e') continue; first[i][j+1]='$'; f=1; break; } } } } for(i=0;i<nont;i++) { printf("\n\nThe first of %c -> ",first[i][0]); for(j=1;first[i][j]!='$';j++) { printf("%c\t",first[i][j]); } } getch(); }
Write a program to calculate follow in grammar in parsing?
#include<iostream.h> #include<conio.h> #include<string.h> int out(char,int n,char []); static char p[10][20]; char c[10]; int k=0; char final[10][20]; void main() { clrscr(); int n,i,j,flag,u,t=4,d=0,l,v; char a[10],first[10][20]; cout<<"\nEnter the no of productions: "; cin>>n; for(i=0;i<n;i++) { cin>>p[i]; } for(j=0;j<n;j++) { k=0; u=0; if(p[j][3]>='A'&&p[j][3]<='Z') { for(i=3;i<strlen(p[j]);i++) { flag=0; k=0; k=out(p[j][i],n,a); for(int l=0;l<k;l++) { if(a[l]=='^') flag=1; else first[j][u++]=a[l]; } if(flag==1) { if(i!=strlen(p[j])-1) { if(p[j][t]>='A'&&p[j][t]<='Z') { t++; continue; } else { first[j][u++]=p[j][t]; break; } } else { first[j][u++]='^'; } } else { break; } } } else { first[j][0]=p[j][3]; first[j][1]='\0'; } } t=0; for(int e=0;e<n;e++) { int flag=1; for(int i=0;i<t;i++) { if(c[i]==p[e][0]) { flag=0; break; } else { flag=1; } } if(flag==1) { c[t]=p[e][0]; d=0; for(int f=0;f<n;f++) { if(p[e][0]==p[f][0]) { for(int o=0;o<strlen(first[f]);o++) { final[t][d++]=first[f][o]; } final[t][d]='\0'; } } t++; } } for(i=0;i<t;i++) cout<<"First of "<<c[i]<<"is : "<<final[i]<<"\n"; //Code to find follow cout<<"\n\nFOLLOW\n"; for(i=0;i<strlen(c);i++) { cout<<"\nFollow of "<<c[i]<<"is :"; if(i==0) cout<<'$'<<" "; for(j=0;j<n;j++) { for(k=3;k<strlen(p[j]);k++) { if(c[i]==p[j][k]) { if(k!=strlen(p[j])-1) { for(l=k+1;l<strlen(p[j]);l++) { flag=0; if(p[j][l]>='A'&&p[j][l]<='Z') { for(u=0;u<strlen(c);u++) { if(c[u]==p[j][l]) break; } for(v=0;v<strlen(c);v++) { if(final[u][v]=='^') flag=1; else cout<<final[u][v]<<" "; } if(flag==1) { if(l==strlen(p[j])-1) cout<<"Follow of "<<p[j][0]<<" "; else continue; } else { break; } } else { cout<<p[j][l]; break; } } } else { cout<<"Follow of "<<p[j][0]<<" "; } } } } } getch(); } int out(char c,int n,char a[]) { for(int i=0;i<n;i++) { if(p[i][0]==c) { if(p[i][3]>='A'&&p[i][3]<='Z') { out(p[i][3],n,a); } else a[k++]=p[i][3]; } } return k; }
