#include<iostream>
using namespace std;
void main() {
int odd,even,number;
odd=0;
even=0;
for(int x=1;x<=10;x=x+1){
cout<<"enter a number\n";
cin>>number;
if(number%2==0){
even++;
}
else {
odd++;
}
}
cout<<"even number has:"<<even;
cout<<"odd number has:"<<odd;
}
for persons who are wondering how to do this its really simple.
/* CLAUDE ANTHONY BURRELL*/
/* C CODE FOR EVENSUM*/
#include<stdio.h>
#include<stdlib.h>
int valChar(int*, char);
int main()
{
int values[11];
char alpha;
int evenSum;
int i;
printf("\t C PROGRAM FOR ALL EVEN NUMBERS AND FIND THE SUM\n\n\n\n");
system("color e");
for(i = 0; i < 10; i++)
{
printf("PLEASE ENTER %d VALUE\n",i); /* READING 10 DIFFERENT NUMBERS IN VALUES*/
scanf("%d",&values[i]);
}//end for
fflush(stdin);
printf("ENTER A CHARACTER\n\n");
scanf("%c",&alpha);
evenSum = valChar(values,alpha);/* RETURNS THE SUM OF EVEN NUMBERS IF THE CHARACTER IS 'A' OR 'a'*/
printf("THE SUM OF THE NUMBER IS = %d ",evenSum);
system("pause");
}//end main
int valChar(int * values, char alpha)
{
int i;
int evenSum = 0;
for(i = 0; i< 10; i++)
{
if (alpha 'A')
{
if (values[i]%2==0)
{
//even = values[i] % 2 = 0;
evenSum = evenSum + values[i];
}//end if
}//end if
else
{
evenSum = evenSum + values[i];
}//end else
}//end for
return evenSum;
}//end valChar
bool is_odd (unsigned n) {
return n & 1;
}
This function simply tests to see if the low-order bit is set using the bitwise-AND binary operator (&). If it is, the number must be odd because the low order bit represents 2^0 which is 1. All other bits represent increasing powers of 2 (2, 4, 8, 16, etc) so the low-order bit alone determines if a number is odd (is set) or even (is not set).
Note that the argument must be unsigned. This is because negative values are implementation-defined and may use ones-complement or twos-complement notation, depending on the implementation. The function will only work with signed integers using twos-complement notation. Twos-complement is more common than ones-complement, however portability is not guaranteed.
For a portable solution that caters for both ones-complement and two-complement signed integers, use the modus operator instead:
bool is_odd (signed n) {
return n % 2;
}
The modus operator returns the remainder after division. When we divide any signed integer by 2, the remainder can only be 0 (is even) or 1 (is odd).
Using while loop, write a program which calculates the product of digits from 1 to 5 and also show these no's vertically.
1. Read in 'n'2. Output n*(n+1)/2> Check the no. in odd or even?Both possible, has no significance.
if (n%2==0) sum=n/2*(n+1); else sum=(n+1)/2*n;
// 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(); }
#include<stdio.h> #include<conio.h> void main() { int a[10],i,evc=0,odc =0;// sum=0; clrscr(); printf("enter 10 no.s in the array\n"); for(i=0;i<=9;i++) { scanf("%d",&a[i]); } for(i=0;i<=9;i++) { if(a[i]%2==0) { evc=evc+1; } else { odc=odc+1; } } printf("Even nos %d\n",evc); printf("odd nos %d\n",odc); getch(); } 0"). This is a perfectly valid and correct approach, but in many implementations, a simple test for the least significant bit ("a[i] & 1") can be more efficient. This also allows to replace logic with arithmetic, which generally is more efficient (i.e. "odc += a[i] & 1")
10
Write a c program to print the 100 to 1 nos
Using while loop, write a program which calculates the product of digits from 1 to 5 and also show these no's vertically.
nos vemos
10,000,000
#include <stdio.h> main() { int n, odd=0, even=0; while (scanf("%d",&n)&&(n!=0)) (n%2)?++odd:++even; printf("odd: %d even: %d\n",odd,even); }
Write (to) us
//sum and product of 3 nos #include #include void main() { int a,b,c; printf("enter the 3 nos"); scanf("%d%d%d",&a,&b,&c); printf("sum of 3 nos",a+b+c); printf("product of 3 nos",a*b*c); getch(); }
even
there cannot be any nos like that.. product of 2 odd nos is odd.. sum of two even nos is even.. that multiplied by six is even too.. subtracting 2 from that also gives an even no.. let x and y be the odd integers. according to the given question xy=6(x+y)-2..here we r actually equatin an odd no and an even no.. which is wrong.. so there cant be any two consecutive odd nos that fit in the question
Each one of them can be expressed as a sum of two primes.
No most are odd nos.