Yes, the Prime Meridian is an imaginary line that divides the Earth into the Western Hemisphere and the Eastern Hemisphere. It runs through Greenwich, England.
THere are two, the Rockies in the West and the Appalachians int the East.
Two thirds of the worlds recorded tornadoes occur int he United States. However, the U.S. portion of tornadoes is likely highly exaggerated as many countries do not keep records on their tornadoes.
Sicily is the largest
// return the larger of any 2 numbers int larger (int a, int b) { return a>b?a:b; } // return the largest of any 3 numbers int largest (int a, int b, int c) { return larger (larger (a, b), c)); } // return the middle value of any 3 numbers int middle (int a, int b, int c) { if (a>b) a^=b^=a^=b; // swap a and b (b is now the larger of the two) if (b>c) b^=c^=b^=c; // swap b and c (c is now the largest of all three) return larger (a, b); // return the larger of a and b }
// assume you want the find the largest of ints a,b,c: int max = (a>b&&a>c?a:b>c?b:c);
Find the largest of two, then find the largest of that value and the third value. int* max (int* a, int* b) { return (a*) > (b*) ? a : b; } int* max_of_three (int* a, int* b, int* c) { return max (max (a, b), c); }
No. Mars is the second smallest planet int he solar system. Only Mercury is smaller.
#include <iostream.h> #include <conio.h> void main() { clrscr(); int largest(int,int,int); cout<<"Enter 3 Integer Numbers\n"; int a,b,c; cin>>a>>b>>c; int result; result=largest(a,b,c); cout<<"\n\nLargest Value of Inputed is "<<result; getch(); } inline largest(int a,int b,int c) { int z; z=(a>b)?((a>c)?a:c):((b>c)?b:c); return(z); }
To find the largest of three numbers, first find the largest of two numbers: int max (int x, int y) { return x<y?y:x; } Now you can use this one function to find the largest of three numbers: int max (int x, int y, int z) { return max (max (x, y), z); }
For this you need two variables, a and b. Generate the first number and assign it to a. Generate the second and assign it to b. If b is greater than a, assign b to a. Generate the third number and assign it to b. If b is greater than a, assign b to a. Print a. A better method is to use the following function to determine the largest of any two values: int max (int x, int y) { return x>y ? x : y; } Once you know the largest of any two values you can easily determine the largest of any three values with a nested call: int max_of_three (int a, int b, int c) { return max (max (a, b), c)); } Note that we determine the largest of a and b first and then pass that value to a second call along with c, thus establishing the largest of all three. To determine the largest of n values, store the numbers in an array and pass the array and its length to this function: int max_of_n (int a[], size_t n) { int m = a[0]; // store first value while (--n) if (a[n]>m) m=a[n]; // update m whenever a[n] is greater return m; }
To find the largest of three numbers you must first find the largest of two numbers: int max (int a, int b) { return a>b?a:b; // or, equivalently: if (a>b) return a; else return b; } Now we can use this function to find the maximum of three: int max3 (int a, int b, int c) { return max (max (a,b), c); }
int lastLargestIndex(int a[],int n) //a=array, n= number of elements in array { int max=a[0],maxp=0; //max=largest no., maxp= position of largest no. for(int i=0;i<n;i++) if(a[i]>=max) { max=a[i]; maxp=i; } return maxp; }
one of the world’s largest radio observatories
int secondmax (int a[], int n){int i, max, second;for (i=0; imax) second= max, max= a[i];else if (i==1 a[i]>second) second= a[1];}return second;}int middle (int a[3]){return secondmax (a, 3);}