45 lines
1.3 KiB
C
45 lines
1.3 KiB
C
/*
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Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
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If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
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For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
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Evaluate the sum of all the amicable numbers under 10000.
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*/
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#include <stdio.h>
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#include <stdlib.h>
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int sumOfDivisors[10001];
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void findSumOfProperDivisors () {
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sumOfDivisors[0] = 0; // placeholder
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for (int i = 1; i <= 10000; i++) {
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int sumOfDivisorsOfI = 1;
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for (int j = 2; j * j <= i; j++) {
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if (j * j == i)
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sumOfDivisorsOfI += j;
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else if (i % j == 0)
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sumOfDivisorsOfI += j + i / j;
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}
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sumOfDivisors[i] = sumOfDivisorsOfI;
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}
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}
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int sumOfAmicable () {
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int sum = 0;
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for (int i = 1; i <= 10000; i++) {
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int value = sumOfDivisors[i];
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if (value <= 10000 && value != i && sumOfDivisors[value] == i) {
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sum += i;
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printf("%d, %d\n", i, value);
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}
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}
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return sum;
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}
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int main (int argc, char* argv[]) {
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findSumOfProperDivisors ();
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printf ("%d\n", sumOfAmicable ());
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}
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