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What are the numerical method of gauss elimination and gauss Jordan method?
Here is the program for Gauss elimination method #include<iostream.h> #include<conio.h> #include<math.h> void main() {float a[6][6],b[6],x[6],t,s; int i,j,n,k; clrscr(); cout<<"Enter the maximum no. of matrix"<<endl; cin>>n; cout<<"Enter th elements of matrix"<<endl; for(i=0;i<n;i++) { for(j=0;j<n;j++) { cin>>a[i][j]; } } cout<<"enter the right constant"<<endl; for(i=0;i<n;i++) { cin>>b[i]; } for(k=0;k<n-1;k++) { for(i=k+1;i<n;i++) { t=a[i][k]/a[k][k]; a[i][k]=0; for(j=k;j<n;j++) { a[i][j]=a[i][j]-(t*a[k][i]); } b[i]=b[i]-(t*b[k]); } } x[n-1]=b[n-1]/a[n-1][n-1]; for(i=n-1;i>=0;i--) { s=0; for(j=i+1;j<n;j++) {s=s+(a[i][j]*x[j]); } x[i]=(b[i]-s)/a[i][i]; } cout<<"the solution is"<<endl; for(i=0;i<n;i++) { cout<<"x["<<i<<"]="<<x[i]<<endl; } getch(); } C program for Gauss Jordan method: #include<iostream.h> #include<conio.h> #include<math.h> void main() {float a[6][6],b[6],x[6],t,s; int i,j,n,k; clrscr(); cout<<"Enter the maximum no. of matrix"<<endl; cin>>n; cout<<"Enter th elements of matrix"<<endl; for(i=0;i<n;i++) { for(j=0;j<n;j++) { cin>>a[i][j]; } } cout<<"enter the right constant"<<endl; for(i=0;i<n;i++) { cin>>a[i][n]; } for(k=0;k<n;k++) { for(i=0;i<n;i++) if(i!=k) { for(j=k+1;j<n+1;j++) { a[i][j]=a[i][j]-(a[i][k]/a[k][k])*a[k][j]); cout<<"the solution is"<<endl; for(i=0;i<n;i++) { x[i]=(a[i][n]/a[i][i]); cout<<"x["<<i<<"]="<<x[i]<<endl; } getch(); }
Write a c program to print mean median and mode?
#include<stdio.h> #define MAXVAL 1000 void sort1(int a[],int n); void median(int a[],int n); void mode(int a[],int n); int main() { int n; int arr[MAXVAL]; int i; printf("Enter the number of elements:"); scanf("%d",&n); printf("Enter the values:"); for(i=0;i<n;i++) { printf("a[%d]=",i); scanf("%d",&arr[i]); } sort1(arr,n); median(arr,n); mode(arr,n); } void sort1(int a[],int n) { int i; int j; int temp; for(i=0;i<n;i++) { for(j=i;j<n;j++) { if(a[i]>a[j]) { temp=a[i]; a[i]=a[j]; a[j]=temp; } } } } void median(int a[],int n) { int median; int mid; if((n%2)==0) { mid=n/2; median=(a[mid-1]+a[mid])/2; } else { mid=(n+1)/2; median=a[mid-1]; } printf("The median is:%d\n",median); } void mode(int a[],int n) { int i; int count1[MAXVAL]; for(i=0;i<n;i++) { count1[i]=0; } for(i=0;i<n;i++) { count1[a[i]]++; } i--; int mode=count1[0]; int j; int k; int flag=0; for(j=0;j<=a[i];j++) { if(count1[j]>count1[mode]) mode=j; } for(j=0;j<=a[i];j++) { for(k=j+1;k<=a[i];k++) { if(count1[j]=count1[k] && count1[j]>count1[mode]) { flag=1; } } } if(flag==1) { printf("Mode cannot be calculated"); } else printf("the Mode is:%d",mode); }
How do you calculate time complexity for insertion sort?
1.for j = 2 to length[A] c1 n 2. do key ¬ A[j] c2 n-1 3. //insert A[j] to sorted sequence A[1..j-1] 0 n-1 4. i ¬ j-1 c4 n-1 5. while i >0 and A[i]>key c5 Sum (j=2->n) tj 6. do A[i+1] ¬ A[i] c6 Sum (j=2->n) (tj -1) 7. i ¬ i-1 c7 Sum (j=2->n) (tj -1) 8. A[i+1] ¬ key c8 n -1 Sum j=2->n tj evaluates to (n(n+1)/2)-1 and j=2->(tj-1) evaluates to n(n-1)/2 thus the highest order term after droping constants becomes n2 thus the complexity is n2
C program to generate the first 50 prime number?
#include <stdio.h> int main () { int n, i, j, k; printf ("2 "); for (i=3,n=2; n<=50; i+=2) { for (j=3,k=1; j<i-1; j+=2) { if (i % j 1) { printf ("%d ", i); n++; } } printf("\n"); return 0; }
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(); }
30 equals J IN THE G N?
30 = Jumps in the Grand National
What is 30 J in the G N?
30 Joules is equivalent to 0.03 gigajoules.
How is 30 J in the G N?
Jumps in the Grand National
How much work is done when a force of 30 n moves an object a distance of 30 m?
The work done is calculated by multiplying the force by the distance the object moves in the direction of the force. So, the work done would be 30 N * 30 m = 900 J.
30 D hath S A J and N?
30 days hath September, April, June, and November.
C program to arrange the set of numbers in descending order?
main (){int i,j,a,n,number[30];printf ("Enter the value of N\n");scanf ("%d", &n);printf ("Enter the numbers \n");for (i=0; i
Program to sort numbers in c plus plus?
/* Bubble sort: code snippet only nos to be sorted are in the array named 'n' of size 'N' for(int i=0;i<N-1;i++) for(int j=i+1;j<N-1-i;j++) if(n[j]>n[j+1]) swap(n[j],n[j+1]); */ /* insertion sort int v,j; for(int i=1;i<N;i++) { v=n[j]; for(int j=i-1;j>0&&n[j]>v;j--) n[j+1]=n[j]; n[j+1]=v; } */
Distance vector routing in c program?
#include<stdio.h> int main() { int n,i,j,k,a[10][10]; printf("\nEnter the number of nodes: "); scanf("%d",&n); for(i=0; i<n; i++) { for(j=0; j<n; j++) { printf("\nEnter the distance between the host %d%d:", i+1,j+1); scanf("%d",&a[i][j]); } } for(i=0; i<n; i++) { for(j=0; j<n; j++) printf("%d\t",a[i][j]); printf("\n"); } for(k=0; k<n; k++) { for(i=0; i<n; i++) { for(j=0; j<n; j++) { if(a[i][j]>a[i][k]+a[k][j]) a[i][j]=a[i][k]+a[k][j]; } } } for(i=0; i<n; i++) { for(j=0; j<n; j++) { b[i][j]=a[i][j]; if(i==j) b[i][j]=0; } } printf("\nThe output matrix is:\n"); for(i=0;i<n;i++) { for(j=0;j<n;j++) printf("%d\t",b[i][j]); printf("\n"); } return 0; }
30 D H S A J and N?
Days have September, April, June, and November
C program to check whether a given matrix is orthogonal or not?
#include<iostream> #include<stdio.h> #include<conio.h> using namespace std; int main() { int a[20][20],b[20][20],c[20][20],i,j,k,m,n,f; cout << "Input row and column of A matrix \n\n"; cin >> n >> m; cout << "\n\nInput A - matrix \n\n"; for(i=0;i<n;++i) for(j=0;j<m;++j) cin >> a[i][j]; cout << "\n\nMatrix A : \n\n"; for(i=0;i<n;++i) { for(j=0;j<m;++j) cout << a[i][j] << " "; cout << "\n\n"; } for(i=0;i<m;++i) for(j=0;j<n;++j) b[i][j]=a[j][i]; cout << "\n\nTranspose of matrix A is : \n\n"; for(i=0;i<m;++i) { for(j=0;j<n;++j) cout << b[i][j] << " "; cout << "\n\n"; } for(i=0;i<m;i++) { for(j=0;j<m;j++){ c[i][j]=0; for(k=0;k<=m;k++) c[i][j]+=a[i][k]*b[k][j]; } } for(i=0;i<m;i++) { for(j=0;j<m;j++) { if((int)c[i][i]==1&&(int)c[i][j]==0) f=1; } } cout<<"\n\n Matrix A * transpose of A \n\n"; for(i=0;i<m;i++) { for(j=0;j<m;j++) cout << c[i][j]; cout << "\n\n"; } if(f==1) cout << "\n\nMatrix A is Orthogonal !!!"; else cout << "\n\nMatrix A is NOT Orthogonal !!!"; getch(); return 0; } -ALOK
What are the next three letters in the following sequnce j f m a m j j a?
these letters stand for the months of the year january february march ... the next letters would be s o n d Hope i helped you lol xx:)
Program for row wise sort for a square matrix in C?
#include<stdio.h> #include<conio.h> void main() { int a[10][10],i,j,k,m,n; printf("enter the order"); scanf("%d",&m,&n); for(i=0;i<m;i++) { for(j=0;j<n;j++) { scanf("%d",&a[i][j]); } } for(k=0;k<m;k++) { for(i=0;i<=m-1;i++) { for(j=0;j<n-i-1;j++) { if(a[k][j]>a[k][j+1]) temp=a[k][j]; a[k][j]=a[k][j+1]; a[k][j+1]=temp; } } } for(i=0;i<m;i++) { for(j=0;j<n;j++) { printf("sorted matrix=%d",a[i][j]); } } getch(); }
