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Write a program that accepts 15 different numbers and find the LCM and HCM?
/* Program to find LCM/HCF of 15 Nos in C++/ C (replace cin with scanf & cout with printf for c) */ /* Developed by - Kishore Kr. Banerjee - papillon_kish@yahoo.com */ #include <iostream.h> #include <conio.h> main() { int num[3],i,j,p,q,tmp,n1,n2,LCM,rem,flag; clrscr(); for(i=0;i<3;i=i+1) { cout<<"Enter No - "<<i+1<<"="; cin>>num[i]; } clrscr(); n1=num[0]; p=n1; for(i=1;i<3;i=i+1) { n2=num[i]; /* Finding HCF */ q=n2; if(p<q) { tmp=q; q=p; p=tmp; } while(p%q!=0) { rem=p%q; q=p; p=rem; if(p<q) { tmp=q; q=p; p=tmp; } } /*finding LCM */ LCM=1; for(j=1;n1%j==0n2%j==0;j=j+1) { if(n1%j==0) { n1=n1/j; flag=1; } if(n2%j==0) { n2=n2/j; flag=1; } if(flag==1) { LCM=LCM*j; } } LCM=LCM*n1*n2; n1=LCM; } cout<<"the LCM ="<<LCM; cout<<"the hcf ="<<q; getch(); } ----- int gcd (int a, int b) { . int tmp; . if (a<0) a= -a; . if (b<0) b= -b; . if (a<b) tmp= a, a= b, b= tmp; . while (b) { . . tmp= a%b; . . a= b; . . b= tmp; . } . return a; } int gcd_n (int n, const int *vect) { . int i, gcdtmp; . for (i=0, gcdtmp=0; i<n && gcdtmp!=1; ++i) . . gcdtmp= gcd (gcdtmp, vect[i]); . return gcdtmp; } int LCM (int a, int b) { . int d= gcd (a, b); . if (d==0) return d; . else return a/d*b; } int lcm_n (int n, const int *vect) { . int i, lcmtmp; . for (i=0, lcmtmp=1; i<n; ++i) . . lcmtmp= LCM (lcmtmp, vect[i]); . return lcmtmp; }
Formulas on Percentage Base and Rate?
P=B×RB=P÷RR=P÷B
How might two probabilities be added?
If A and B are two events then P(A or B) = P(A) + P(B) - P(A and B)
Top down parser source code in c plus plus?
#include<iostream.h> #include<conio.h> #include<string.h> class parse { int nt,t,m[20][20],i,s,n,p1,q,k,j; char p[30][30],n1[20],t1[20],ch,b,c,f[30][30],fl[30][30]; public: int scant(char); int scannt(char); void process(); void input(); }; int parse::scannt(char a) { int c=-1,i; for(i=0;i<nt;i++) { if(n1[i]==a) { return i; } } return c; } int parse::scant(char b) { int c1=-1,j; for(j=0;j<t;j++) { if(t1[j]==b) { return j; } } return c1; } void parse::input() { cout<<"Enter the number of productions:"; cin>>n; cout<<"Enter the productions one by one"<<endl; for(i=0;i<n;i++) cin>>p[i]; nt=0; t=0; } void parse::process() { for(i=0;i<n;i++) { if(scannt(p[i][0])==-1) n1[nt++]=p[i][0]; } for(i=0;i<n;i++) { for(j=3;j<strlen(p[i]);j++) { if(p[i][j]!='e') { if(scannt(p[i][j])==-1) { if((scant(p[i][j]))==-1) t1[t++]=p[i][j]; } } } } t1[t++]='$'; for(i=0;i<nt;i++) { for(j=0;j<t;j++) m[i][j]=-1; } for(i=0;i<nt;i++) { cout<<"Enter first["<<n1[i]<<"]:"; cin>>f[i]; } for(i=0;i<nt;i++) { cout<<"Enter follow["<<n1[i]<<"]:"; cin>>fl[i]; } for(i=0;i<n;i++) { p1=scannt(p[i][0]); if((q=scant(p[i][3]))!=-1) m[p1][q]=i; if((q=scannt(p[i][3]))!=-1) { for(j=0;j<strlen(f[q]);j++) m[p1][scant(f[q][j])]=i; } if(p[i][3]=='e') { for(j=0;j<strlen(fl[p1]);j++) m[p1][scant(fl[p1][j])]=i; } } for(i=0;i<t;i++) cout<<"\t"<<t1[i]; cout<<endl; for(j=0;j<nt;j++) { cout<<n1[j]; for(i=0;i<t;i++) { cout<<"\t"<<" "; if(m[j][i]!=-1) cout<<p[m[j][i]]; } cout<<endl; } } void main() { clrscr(); parse p; p.input(); p.process(); getch(); }
What is the Formula for odds for either of two events?
If A and B are mutually exclusive event then Probability of A or B is P(A)+P(B). If they are not mutually exclusive then it is that minus the probability of the P(A)+P(B) That is to say P( A or B)= P(A)+P(B)- P(A and B). Of course it is clear that if they are mutually exclusive, P(A and B)=0 and we have the first formula.
