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Project Euler solving Problem 01

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koseokkyu 2017-11-27 17:00:21 +09:00
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19
Project Euler/Problem 01/sol1.c Normal file
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/*
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
*/
#include <stdio.h>
int main() {
int n = 0;
int sum = 0;
scanf("%d", &n);
for (int a = 0; a < n; a++) {
if ((a % 3 == 0) || (a % 5 == 0)) {
sum += a;
}
}
printf("%d\n", sum);
}

24
Project Euler/Problem 01/sol2.c Normal file
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/*
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
'''
'''
This solution is based on the pattern that the successive numbers in the series follow: 0+3,+2,+1,+3,+1,+2,+3.
*/
#include <stdio.h>
int main() {
int n = 0;
int sum = 0;
scanf("%d", &n);
int terms = (n - 1) / 3;
sum += ((terms)*(6 + (terms - 1) * 3)) / 2; //sum of an A.P.
terms = (n - 1) / 5;
sum += ((terms)*(10 + (terms - 1) * 5)) / 2;
terms = (n - 1) / 15;
sum -= ((terms)*(30 + (terms - 1) * 15)) / 2;
printf("%d\n", sum);
}

49
Project Euler/Problem 01/sol3.c Normal file
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/*
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
'''
'''
This solution is based on the pattern that the successive numbers in the series follow: 0+3,+2,+1,+3,+1,+2,+3.
*/
#include <stdio.h>
int main() {
int n = 0;
int sum = 0;
int num = 0;
scanf("%d", &n);
while (1) {
num += 3;
if (num >= n)
break;
sum += num;
num += 2;
if (num >= n)
break;
sum += num;
num += 1;
if (num >= n)
break;
sum += num;
num += 3;
if (num >= n)
break;
sum += num;
num += 1;
if (num >= n)
break;
sum += num;
num += 2;
if (num >= n)
break;
sum += num;
num += 3;
if (num >= n)
break;
sum += num;
}
printf("%d\n", sum);
}