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72 lines
2.0 KiB
C
72 lines
2.0 KiB
C
//ProjectEuler/C/Problem1.c
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//Matthew Ellison
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// Created: 03-08-19
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//Modified: 03-28-19
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//What is the sum of all the multiples of 3 or 5 that are less than 1000
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//Unless otherwise listed all non-standard includes are my own creation and available from https://bibucket.org/Mattrixwv/myHelpers
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/*
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Copyright (C) 2019 Matthew Ellison
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This program is free software: you can redistribute it and/or modify
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it under the terms of the GNU Lesser General Public License as published by
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the Free Software Foundation, either version 3 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public License
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along with this program. If not, see <https://www.gnu.org/licenses/>.
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*/
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#include <stdio.h>
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#include <inttypes.h>
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#include "Stopwatch.h"
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const int MAX_NUMBER = 1000;
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int main(){
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//Setup the stopwatch
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struct Stopwatch timer;
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initStopwatch(&timer);
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//Setup the variables you will need
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uint64_t fullSum = 0; //For the sum of all the numbers
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//Start the timer
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startStopwatch(&timer);
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//Step through ever number < 1000 and see if either 3 or 5 divides it evenly
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for(uint64_t cnt = 1;cnt < MAX_NUMBER;++cnt){
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if((cnt % 3) == 0){
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fullSum += cnt;
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}
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else if((cnt % 5) == 0){
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fullSum += cnt;
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}
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}
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//Stop the timer
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stopStopwatch(&timer);
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char* timerStr = getStrStopwatch(&timer);
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//Print the results
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printf("The sum of all the numbers < %d that are divisible by 3 or 5 is %lu\n", MAX_NUMBER, fullSum);
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printf("It took %s to run this algorithm\n", timerStr);
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//Free the memory taken up by the string
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free(timerStr);
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timerStr = NULL;
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return 0;
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}
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/* Results:
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The sum of all the numbers < 1000 that are divisible by 3 or 5 is 233168
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It took 3.000 microseconds to run this algorithm
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*/
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