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#include<stdio.h> #include<conio.h> void main() { int n; printf("Enter any number"); scanf("%d",&n); sum=((n*(n+1))/2); printf("The sum of first 100 natural numbers is\t%d",sum); }

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15y ago
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7y ago



bool is_prime (int num) {

int factor, max_factor;

if (num<0) n*= (-1); // make num positive

if (num<2) return false; // 2 is the lowest positive prime

if ((num%2)==0) return num==2; // 2 is the only even prime

max_factor = sqrt (num) + 1; // if no factors less than max_factor, then none greater either

for (factor=3; factor<=max_factor; factor+=2) // test all odd factors in half-closed range [3:max_factor)

if ((num%factor)==0) return false; // num is composite

return true; // num has no factors, so it must be prime


int main (void)


const int min = 100;

const int max = 500;

int sum=0;

for (int num=min; num<=max; ++num) if (is_prime (num)) sum+=num;

printf ("Sum of all prime numbers from %d to %d is %d\n", min, max, sum);

return 0;


EDIT: Previous answer produced correct result but was highly inefficient. The modified code above is more efficient but could be improved by ignoring all even values, thus the range becomes 101 through 499, inclusive, and increments in steps of 2 instead of 1 (e.g., num += 2 instead of ++num).

The is_prime(n) algorithm has a worst-case time complexity of O(sqrt(n)/2) with a constant space complexity O(1). While adequate for this otherwise trivial program, the algorithm would need to be improved to cope with substantially larger values. A lookup table is one solution since all prime numbers are constant variables and, if the table is saved to disk, would only need to be generated once. A binary search of the table can be achieved in O(log n) time, you simply load the table into an array of appropriate size as and when it is needed.

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14y ago

#include <stdio.h>



int i;

for (i=0;i<100;++i)



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Q: Write C program to find the sum of first 100 numbers?
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