30 lines
815 B
C
30 lines
815 B
C
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#include <stdio.h>
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/*
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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.
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Find the sum of all the multiples of 3 or 5 below 1000.
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https://projecteuler.net/problem=1
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*/
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int main(int argc,char**argv) {
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int counter1=0;
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int counter2=0;
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int sum=0;
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while (counter1<1000||counter2<1000) {
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counter1+=3;
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if (counter1!=counter2&&counter1<1000) {
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sum+=counter1;
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printf("Sum is %d (+%d) [3]\n",sum,counter1);
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}
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if (counter1>counter2) {
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counter2+=5;
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if (counter1!=counter2&&counter2<1000) {
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sum+=counter2;
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printf("Sum is %d (+%d) [5]\n",sum,counter2);
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}
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}
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}
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printf("%d",sum);
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return 0;
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}
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