mirror of https://github.com/TheAlgorithms/C
Project Euler solving Problem 01
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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 1000.
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*/
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#include <stdio.h>
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int main() {
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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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for (int a = 0; a < n; a++) {
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if ((a % 3 == 0) || (a % 5 == 0)) {
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sum += a;
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}
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}
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printf("%d\n", sum);
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}
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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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'''
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This solution is based on the pattern that the successive numbers in the 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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int main() {
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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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}
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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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'''
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This solution is based on the pattern that the successive numbers in the 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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int main() {
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int n = 0;
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int sum = 0;
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int num = 0;
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scanf("%d", &n);
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while (1) {
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num += 3;
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if (num >= n)
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break;
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sum += num;
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num += 2;
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if (num >= n)
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break;
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sum += num;
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num += 1;
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if (num >= n)
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break;
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sum += num;
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num += 3;
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if (num >= n)
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break;
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sum += num;
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num += 1;
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if (num >= n)
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break;
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sum += num;
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num += 2;
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if (num >= n)
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break;
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sum += num;
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num += 3;
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if (num >= n)
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break;
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sum += num;
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}
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printf("%d\n", sum);
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}
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