2.2 KiB
2.2 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f3b61000cf542c50fec8 | Problem 73: Counting fractions in a range | 5 | 302186 | problem-73-counting-fractions-in-a-range |
--description--
Consider the fraction, \frac{n}{d}
, where n
and d
are positive integers. If n
< d
and highest common factor, {HCF}(n, d) = 1
, it is called a reduced proper fraction.
If we list the set of reduced proper fractions for d
≤ 8 in ascending order of size, we get:
\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \mathbf{\frac{3}{8}, \frac{2}{5}, \frac{3}{7}}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}
It can be seen that there are 3
fractions between \frac{1}{3}
and \frac{1}{2}
.
How many fractions lie between \frac{1}{3}
and \frac{1}{2}
in the sorted set of reduced proper fractions for d
≤ limit
?
--hints--
countingFractionsInARange(8)
should return a number.
assert(typeof countingFractionsInARange(8) === 'number');
countingFractionsInARange(8)
should return 3
.
assert.strictEqual(countingFractionsInARange(8), 3);
countingFractionsInARange(1000)
should return 50695
.
assert.strictEqual(countingFractionsInARange(1000), 50695);
countingFractionsInARange(6000)
should return 1823861
.
assert.strictEqual(countingFractionsInARange(6000), 1823861);
countingFractionsInARange(12000)
should return 7295372
.
assert.strictEqual(countingFractionsInARange(12000), 7295372);
--seed--
--seed-contents--
function countingFractionsInARange(limit) {
return true;
}
countingFractionsInARange(8);
--solutions--
function countingFractionsInARange(limit) {
let result = 0;
const stack = [[3, 2]];
while (stack.length > 0) {
const [startDenominator, endDenominator] = stack.pop();
const curDenominator = startDenominator + endDenominator;
if (curDenominator <= limit) {
result++;
stack.push([startDenominator, curDenominator]);
stack.push([curDenominator, endDenominator]);
}
}
return result;
}