it's simple:
{
int a,b,c,d;
printf("enter the numbers");
scanf("%d %d %d" &a &b &c);
d=a+b+c;
printf("the sum is=%d" d);
}
simple program
algorithm is a way to solve your problem
Three ways.. Multiply n by itself. Calculate Sum[2i+1,{i,0,n-1}] Calculate Sum[n,{i,1,n}]
public static final int getSum(final int n) { int sum = 0; for(int i = 1; i <= n; ++i) { sum += i; } return sum; }
In QBASIC, you can write a simple program to input the number 64751315 and sum its digits as follows: DIM sum AS INTEGER sum = 0 INPUT "Enter a number: "; number FOR i = 1 TO LEN(number) sum = sum + VAL(MID$(number, i, 1)) NEXT PRINT "The sum of the digits is "; sum This program prompts the user to input a number, iterates through each digit, converts it to an integer, and adds it to the total sum, which is then printed out.
The quotient of the sum of a number and three, and four is seven
To calculate the sum of all even numbers starting from 20 until the sum exceeds 1000, you can initialize a variable for the sum and a counter starting at 20. In a loop, add the counter to the sum and increment the counter by 2 (to keep it even) until the sum exceeds 1000. The final sum will be the total of all even numbers added. Here's a simple pseudocode example: sum = 0 number = 20 while sum <= 1000: sum += number number += 2
The sum of four and the product of three and a number xxxx.
n+3 the sum of a number and three means a number plus three.
int sum = 0; int n = 0; while( sum <= 999 ) { sum += (++n); }
Their sum is three times the middle number.
There are not three odd primes with the sum of 14. The sum of three odd primes will be an odd number.