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Results 1 - 7 of 7 for gCD (0.07 sec)
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android/guava-tests/test/com/google/common/math/IntMathTest.java
for (int b : POSITIVE_INTEGER_CANDIDATES) { assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(IntMath.gcd(a, b))); } } } public void testGCDZero() { for (int a : POSITIVE_INTEGER_CANDIDATES) { assertEquals(a, IntMath.gcd(a, 0)); assertEquals(a, IntMath.gcd(0, a)); } assertEquals(0, IntMath.gcd(0, 0)); } public void testGCDNegativePositiveThrows() {
Registered: Fri Nov 01 12:43:10 UTC 2024 - Last Modified: Sat Oct 19 00:26:48 UTC 2024 - 23.1K bytes - Viewed (0) -
guava-tests/test/com/google/common/math/IntMathTest.java
for (int b : POSITIVE_INTEGER_CANDIDATES) { assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(IntMath.gcd(a, b))); } } } public void testGCDZero() { for (int a : POSITIVE_INTEGER_CANDIDATES) { assertEquals(a, IntMath.gcd(a, 0)); assertEquals(a, IntMath.gcd(0, a)); } assertEquals(0, IntMath.gcd(0, 0)); } public void testGCDNegativePositiveThrows() {
Registered: Fri Nov 01 12:43:10 UTC 2024 - Last Modified: Sat Oct 19 00:26:48 UTC 2024 - 23.1K bytes - Viewed (0) -
guava-tests/test/com/google/common/math/LongMathTest.java
assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(LongMath.gcd(a, b))); } } } @GwtIncompatible // TODO public void testGCDZero() { for (long a : POSITIVE_LONG_CANDIDATES) { assertEquals(a, LongMath.gcd(a, 0)); assertEquals(a, LongMath.gcd(0, a)); } assertEquals(0, LongMath.gcd(0, 0)); } @GwtIncompatible // TODO
Registered: Fri Nov 01 12:43:10 UTC 2024 - Last Modified: Fri Oct 18 15:00:32 UTC 2024 - 30.6K bytes - Viewed (0) -
android/guava/src/com/google/common/math/LongMath.java
int bTwos = Long.numberOfTrailingZeros(b); b >>= bTwos; // divide out all 2s while (a != b) { // both a, b are odd // The key to the binary GCD algorithm is as follows: // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b). // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two. // We bend over backwards to avoid branching, adapting a technique from
Registered: Fri Nov 01 12:43:10 UTC 2024 - Last Modified: Wed Oct 09 16:39:37 UTC 2024 - 45.2K bytes - Viewed (0) -
guava/src/com/google/common/math/LongMath.java
int bTwos = Long.numberOfTrailingZeros(b); b >>= bTwos; // divide out all 2s while (a != b) { // both a, b are odd // The key to the binary GCD algorithm is as follows: // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b). // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two. // We bend over backwards to avoid branching, adapting a technique from
Registered: Fri Nov 01 12:43:10 UTC 2024 - Last Modified: Wed Oct 09 16:39:37 UTC 2024 - 45.2K bytes - Viewed (0) -
cmd/endpoint-ellipses.go
// all the ellipses sizes. func getDivisibleSize(totalSizes []uint64) (result uint64) { gcd := func(x, y uint64) uint64 { for y != 0 { x, y = y, x%y } return x } result = totalSizes[0] for i := 1; i < len(totalSizes); i++ { result = gcd(result, totalSizes[i]) } return result } // isValidSetSize - checks whether given count is a valid set size for erasure coding.
Registered: Sun Nov 03 19:28:11 UTC 2024 - Last Modified: Wed Aug 14 17:11:51 UTC 2024 - 14.7K bytes - Viewed (0) -
android/guava/src/com/google/common/primitives/Ints.java
// moved at that point. Otherwise, we can rotate the cycle a[1], a[1 + d], a[1 + 2d], etc, // then a[2] etc, and so on until we have rotated all elements. There are gcd(d, n) cycles // in all. // (3) "Successive". We can consider that we are exchanging a block of size d (a[0..d-1]) with a // block of size n-d (a[d..n-1]), where in general these blocks have different sizes. If we
Registered: Fri Nov 01 12:43:10 UTC 2024 - Last Modified: Fri Oct 25 18:05:56 UTC 2024 - 31K bytes - Viewed (0)