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C program for optimal merge pattern?
#include<iostream.h> #include<conio.h> void main() { clrscr(); int i,k,a[10],c[10],n,l; cout<<"Enter the no. of elements\t"; cin>>n; cout<<"\nEnter the sorted elments for optimal merge pattern"; for(i=0;i<n;i++) { cout<<"\t"; cin>>a[i]; } i=0;k=0; c[k]=a[i]+a[i+1]; i=2; while(i<n) { k++; if((c[k-1]+a[i])<=(a[i]+a[i+1])) { c[k]=c[k-1]+a[i]; } else { c[k]=a[i]+a[i+1]; i=i+2; while(i<n) { k++; if((c[k-1]+a[i])<=(c[k-2]+a[i])) { c[k]=c[k-1]+a[i]; } else { c[k]=c[k-2]+a[i]; }i++; } }i++; } k++; c[k]=c[k-1]+c[k-2]; cout<<"\n\nThe optimal sum are as follows......\n\n"; for(k=0;k<n-1;k++) { cout<<c[k]<<"\t"; } l=0; for(k=0;k<n-1;k++) { l=l+c[k]; } cout<<"\n\n The external path length is ......"<<l; getch(); }
How many ways can I select a committee of 3 people from 10 people?
To select a committee of 3 people from 10, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ). Here, ( n = 10 ) and ( k = 3 ). This gives ( C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 ). Therefore, there are 120 ways to select a committee of 3 people from 10.
Write a c programm for pyramid of given character?
#include<stdio.h> main() { int i,j,k,n; char c; printf("enter the # of rows of graphical output\n"); scanf("%d",&n); printf("enter the character you want to print\n"); scanf("%c",&c); for(i=1;i<=n;i++) { for (k=1;k<=(n-i);k++) { printf(" "); } for(j=0;j<i;j++) { printf("%c",c); printf(" "); } for(k=1;k<=(n-i-1);k++) { printf(" "); } printf("\n"); } getch(); }
How do you make a diamond using for loop in C plus plus?
Here is the code to do it: #include<stdio.h> main() { int n, c, k, space = 1; //Here we ask for the number of rows would be : printf("Enter number of rows\n"); scanf("%d",&n); space = n - 1; //This is the first half of the diamond for ( k = 1 ; k <= n ; k++ ) { for ( c = 1 ; c <= space ; c++ ) printf(" "); space--; for ( c = 1 ; c <= 2*k-1 ; c++) printf("*"); printf("\n"); } space = 1; //Here is the second half of the diamond for ( k = 1 ; k <= n - 1 ; k++ ) { for ( c = 1 ; c <= space; c++) printf(" "); space++; for ( c = 1 ; c <= 2*(n-k)-1 ; c++ ) printf("*"); printf("\n"); } return 0; } Hope that helped :)
Least common multiple in C programming?
The following code for example is a solution (you could do it with less variables, but this is more readable):int GCD(int a, int b){int n, k, c;n = (a>b)?a:b;k = (a>b)?b:a;while (k){c = n%k;n=k;k=c;}return n;}
Wap to merge two arrays using c?
#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]); }
How many ways can an IRS auditor select 5 of 10 tax returns for an audit?
The number of ways an IRS auditor can select 5 tax returns from a total of 10 can be calculated using the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ), where ( n ) is the total number of items, and ( k ) is the number of items to choose. In this case, it would be ( C(10, 5) = \frac{10!}{5!5!} = 252 ). Thus, there are 252 different ways for the auditor to select 5 tax returns for an audit.
What is the value for Coulombs constant k?
8.998 X 10^9 N*m^2/C^2
What is the matlab code to generate comb function?
In MATLAB, you can generate the comb function (also known as the comb or binomial coefficient function) using the nchoosek function. Here is a simple code snippet to create a comb function: function C = comb(n, k) C = nchoosek(n, k); end You can call this function by passing the values of n and k to get the binomial coefficient ( C(n, k) ). For example, comb(5, 2) would return 10.
C program for displaying perfect numbers using nested loop?
for (n=1; n<1000; ++n) { for (sum=0, k=1; k<=n/2; ++k) if (n%k==0) sum += k; if (sum==n) printf ("%d\n", n); }
Can you unscramble the letters C U T K N O C K O?
knockout
Program in c finding average of two and three digits nos?
#include<stdio.h> #include<conio.h> void main() { int i,j,k,a[10],n; clrscr(); scanf("%d",&n); for(i=0;i<3;i++) { n=n%10; j=0; a[j]=n; j++; } for(j=0;j>3;j++) { a[k]=0; a[k]=a[k]+a[j]; avg=(a[k]/3); printf("the average value of digit will be %d",avg); } getch(); } }
