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https://github.com/TheAlgorithms/C
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29 lines
784 B
C
29 lines
784 B
C
/**
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* \file
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* \brief [Problem 1](https://projecteuler.net/problem=1) solution
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*
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* If we list all the natural numbers below 10 that are multiples of 3 or 5,
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* we get 3,5,6 and 9. The sum of these multiples is 23.
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* Find the sum of all the multiples of 3 or 5 below N.
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*
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* This solution is based on the pattern that the successive numbers in the
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* series follow: 0+3,+2,+1,+3,+1,+2,+3.
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*/
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#include <stdio.h>
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/** Main function */
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int main()
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{
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int n = 0;
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int sum = 0;
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scanf("%d", &n);
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int terms = (n - 1) / 3;
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sum += ((terms) * (6 + (terms - 1) * 3)) / 2; // sum of an A.P.
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terms = (n - 1) / 5;
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sum += ((terms) * (10 + (terms - 1) * 5)) / 2;
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terms = (n - 1) / 15;
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sum -= ((terms) * (30 + (terms - 1) * 15)) / 2;
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printf("%d\n", sum);
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} |