(defun max3 (a b c)
(cond ((> a b) (cond ((> a c) a) (t c)))
((> b c) b)
(t c)
)
)
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// HI THIS IS MAYANK PATEL /*C Program to find Maximum of 3 nos. using Nested if*/ #include<stdio.h> #include<conio.h> void main() { int a,b,c; // clrscr(); printf("Enter three number\n\n"); scanf("d%d",&a,&b,&c); if(a>b) { if(a>c) { printf("\n a is maximum"); } else { printf("\n c is maximum"); } } else { if(b>c) { printf("\n b is maximum"); } else { printf("\n c is maximum"); } } getch(); }
Cls input "enter two no.s ",a,b sum=a+b print "sum = ";sum end
// largest = largest of a, b, c public class largest { public static void main(String args[]) { int a,b,c,largest; a=0; b=0; c=0; a=Integer.parseInt(args[0]); b=Integer.parseInt(args[1]); c=Integer.parseInt(args[2]); largest=a>b?(a>c?a:c):(b>c?b:c); System.out.println("The largest no. of "+a+","+b+"and"+c+"is"+largest); } }
You write it exactly the same as you would write it in any other verions of C++, by taking user input to determine the three sides of your triangle. In other words, input three real numbers. What you do with those three numbers is entirely up to you, but presumably you'd want to calculate the angles of a triangle given the length of its three sides. For that you would need to use the cosine rule which states that for any triangle with angles A, B and C whose opposing sides are a, b and c respectively, cos A = (b2 + c2 - a2)/2bc and cos B = (c2 + a2 - b2)/2ca. Knowing two angles, A and B, you can easily work out that angle C must be 180 - (A + B).
#include<stdio.h> #include<conio.h> void main() { int a,b,c; int Result; printf("enter the value of a:"); scanf("%d", &a); printf("enter the value of b"); scanf("%d", &b); printf("enter the value of c"); scanf("%d", &c); Result=a*b*c; printf("%d", Result); getch(); }