/**
* Store partial product result lines in the Calculator output lines.
*/
func (calc *Calculator) setPartialLines() {
var (
opDigits = calc.operands[1].String()
opLength = len(opDigits)
)
// From largest to smallest place values (i.e., last to first partial result
// lines)
for index, digit := range opDigits {
var (
placeValue big.Int
buffer bytes.Buffer
// Amount of right indentation this line has and also the position
// relative to the first partial result line (right after the first
// ruler)
offset = opLength - index - 1
)
placeValue.SetString(fmt.Sprintf("%c", digit), 10)
placeValue.Mul(&placeValue, &calc.operands[0])
// We're right-padding with spaces, but these will be stripped out later
buffer.WriteString(placeValue.String())
buffer.WriteString(strings.Repeat(" ", offset))
calc.lines[3+offset] = buffer.String()
}
}
func fibo(n *big.Int) *big.Int {
s := n.String()
if r, ok := fiboCache[s]; ok {
return r
}
i0 := big.NewInt(0)
if n.Cmp(i0) <= 0 {
return i0
}
if n.Cmp(big.NewInt(2)) <= 0 {
return big.NewInt(1)
}
n1 := big.NewInt(-1)
n1.Add(n1, n)
n1 = fibo(n1)
n2 := big.NewInt(-2)
n2.Add(n2, n)
n2 = fibo(n2)
i0.Add(n1, n2)
fiboCache[s] = i0
return i0
}
func digitSum(num *big.Int) (sum int) {
sum = 0
str := num.String()
for _, v := range str {
sum += v - 48
}
return
}
func info(a *big.Int, n uint) {
dtrunc := int64(float64(a.BitLen())*.30103) - 10
var first, rest big.Int
rest.Exp(first.SetInt64(10), rest.SetInt64(dtrunc), nil)
first.Quo(a, &rest)
fstr := first.String()
fmt.Printf("%d! begins %s... and has %d digits.\n",
n, fstr, int64(len(fstr))+dtrunc)
}
func sumDigits(in *big.Int) (out int) {
s := in.String()
for _, c := range s {
out = out + (c - '0')
}
return
}
// printNumber outputs the given big.Int and also appends a ".5" if there is an
// apple that was divided in half.
func (calc *Calculator) printNumber(number *big.Int) {
fmt.Print(number.String())
if calc.odd {
fmt.Print(".5")
}
fmt.Println()
}
func SaveKeys(n, e, d, p, q *big.Int) {
pu := strings.Bytes(n.String() + "\n" + e.String() + "\n")
pr := strings.Bytes(p.String() + "\n" + q.String() + "\n" + d.String() + "\n")
if ioutil.WriteFile(*publicKeyFile, pu, 0600) != nil ||
ioutil.WriteFile(*privateKeyFile, pr, 0600) != nil {
panic("Writing problems")
}
}
/**
* Store the solution in the Calculator output lines.
*/
func (calc *Calculator) setSolutionLine() {
var solution big.Int
if calc.operator == '+' {
solution.Add(&calc.operands[0], &calc.operands[1])
} else if calc.operator == '-' {
solution.Sub(&calc.operands[0], &calc.operands[1])
} else {
solution.Mul(&calc.operands[0], &calc.operands[1])
}
calc.lines[len(calc.lines)-1] = solution.String()
}
func main() {
const limit = 1000
fibonacci := make_fibonacci()
var last_fib *big.Int
var last_idx int
for i := 1; ; i++ {
last_fib = fibonacci()
if len(last_fib.String()) >= limit {
last_idx = i
break
}
}
fmt.Printf("Fib(%d) = %v, is the first term in the Fibonacci sequence >= %v\n", last_idx, last_fib, limit)
}
func FindLastNonZeroDigit(val *big.Int) int {
sval := val.String()
slen := len(sval)
fmt.Println("Len: ", slen)
lastNonZero := slen
for i := slen - 1; i >= 0; i-- {
c := string(sval[i])
if c != string("0") {
fmt.Println("i: ", i)
lastNonZero = i
break
}
}
return lastNonZero
}
/**
* Find the next palindrome after the current number.
*/
func (factory *PalinFactory) Next() string {
var (
buffer bytes.Buffer
numberLen = len(factory.number)
oddLength = numberLen%2 > 0
// Greedily split the number into two halves. Greedily meaning for
// odd-length numbers, the middle digit belongs to both sides (e.g.,
// "987" splits into "98" and "87").
leftSide = factory.number[0 : (numberLen+1)/2]
rightSide = factory.number[numberLen/2:]
mirroredSide = Reverse(leftSide)
leftLength = len(leftSide)
)
// Before we create a palindrome, we can determine if the resulting
// palindrome will not be greater than the original number by mirroring the
// left side and comparing it to the right side.
//
// Using 123456 as an example, the palindrome will be 123321, which is less
// than 123456. If we mirror the left side, we get 321. The right side is
// 456. Since 321 <= 456, we know the palindrome will not be greater than
// the original number.
if !Greater(mirroredSide, rightSide) {
// Since it is not greater, we can increment the palindrome by
// incrementing the left side. This is essentially incrementing the
// middle digits of the palindrome, which results in the very next
// palindrome. Since we are also incrementing the left side, the
// resulting palindrome will also be greater than the original number.
leftSide = Increment(leftSide)
// One obvious optimization is to set leftSide here and then set it
// again in the next inner if {}, rather than calling leftValue.String()
// twice, but that seems to consistently be slower for some reason.
// If we introduced a new digit, this changes the evenness/oddness of
// the number, and we need to account for this when constructing the
// palindrome. Since we greedily split the number, going from odd to
// even means we need to drop a digit on the left.
//
// For example, the left side of 999 is "99". The left side of 1001 is
// "10". Notice both have two digits. However, if we increment "99"
// (as a left side), we get "100", so we need to divide by 10 to get the
// two digits that will be mirrored to form the palindrome 1001. This
// doesn't need to be done when going from odd to even, because the left
// side has one more digit (left side of 99 is "9", left side of 101 is
// "10").
if len(leftSide) != leftLength {
oddLength = !oddLength
if !oddLength {
var leftValue big.Int
leftValue.SetString(leftSide, 10)
leftValue.Div(&leftValue, big.NewInt(10))
leftSide = leftValue.String()
}
}
mirroredSide = Reverse(leftSide)
}
buffer.WriteString(leftSide)
// Mirror the left side onto the right to form a palindrome
if oddLength {
// When odd, the left and right sides "overlap" on the middle digit, so
// don't include in on the right
buffer.WriteString(mirroredSide[1:])
} else {
buffer.WriteString(mirroredSide)
}
return buffer.String()
}