


Translate the following content into Chinese: In C programming, find the sum of numbers within N that can be divided by 2 or 5
The sum of n natural numbers that can be divided by 2 or 5 can be found by finding the sum of all natural numbers within N that can be divided by 2 and the sum of all natural numbers that can be divided by 5 within N. Come and find out. Subtract these two sums by the sum of natural numbers within N that are divisible by 10, and that's what we want. This method is an efficient way to find the sum of large values of n.
Some of you must be thinking of using loops and conditional statements and then adding up all numbers divisible by 2 or 5, but this approach is inefficient as it has time complexity of order n . This means that for larger values of n, the program will run the loop n times. And doing it this way makes the program heavier.
Find the formula for the sum of n natural numbers that can be divisible by 2
Sum2 = ((n / 2) * (4 + (n / 2 - 1) * 2)) / 2
Find the formula for the sum of n natural numbers that can be divisible by 5
Sum5 = ((n / 5) * (10 + (n / 5 - 1) * 5)) / 2
Find the formula for the sum of n natural numbers that can be divisible by 5 Formula for the sum of natural numbers divisible by 10
Sum10 = ((n / 10) * (20 + (n / 10 - 1) * 10)) / 2
Expected output
Sum = Sum2 + Sum5 - Sum10
Example
#include <stdio.h> int main() { int n = 25; long int sum2, sum5, sum10; sum2 = ((n / 2) * (4 + (n / 2 - 1) * 2)) / 2; sum5 = ((n / 5) * (10 + (n / 5 - 1) * 5)) / 2; sum10 = ((n / 10) * (20 + (n / 10 - 1) * 10)) / 2; long int sum = sum2 + sum5 - sum10; printf("Sum is %d", sum); return 0; }
Output
Sum is 201
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