英文:
Calculating large exponentiation in Golang
问题
我一直在尝试在Golang中计算2^100
。我了解到了数字类型的限制,并尝试使用math/big
包。以下是我尝试过的代码,但我无法弄清楚为什么它不起作用。
我使用了二的幂次计算的方法来计算指数。
package main
import (
"fmt"
"math/big"
)
func main() {
two := big.NewInt(2)
hundred := big.NewInt(50)
fmt.Printf("2 ** 100 is %d\n", ExpByPowOfTwo(two, hundred))
}
func ExpByPowOfTwo(base, power *big.Int) *big.Int {
result := big.NewInt(1)
zero := big.NewInt(0)
for power != zero {
if modBy2(power) != zero {
multiply(result, base)
}
power = divideBy2(power)
base = multiply(base, base)
}
return result
}
func modBy2(x *big.Int) *big.Int {
return big.NewInt(0).Mod(x, big.NewInt(2))
}
func divideBy2(x *big.Int) *big.Int {
return big.NewInt(0).Div(x, big.NewInt(2))
}
func multiply(x, y *big.Int) *big.Int {
return big.NewInt(0).Mul(x, y)
}
英文:
I've been trying to calculating 2^100
in Golang. I understand the limit of numeric type and tried using math/big
package. Here's what I've tried but I can't figure out why it doesn't work.
I've used computation by powers of two method to calculate the exponentiation.
package main
import (
"fmt"
"math/big"
)
func main() {
two := big.NewInt(2)
hundred := big.NewInt(50)
fmt.Printf("2 ** 100 is %d\n", ExpByPowOfTwo(two, hundred))
}
func ExpByPowOfTwo(base, power *big.Int) *big.Int {
result := big.NewInt(1)
zero := big.NewInt(0)
for power != zero {
if modBy2(power) != zero {
multiply(result, base)
}
power = divideBy2(power)
base = multiply(base, base)
}
return result
}
func modBy2(x *big.Int) *big.Int {
return big.NewInt(0).Mod(x, big.NewInt(2))
}
func divideBy2(x *big.Int) *big.Int {
return big.NewInt(0).Div(x, big.NewInt(2))
}
func multiply(x, y *big.Int) *big.Int {
return big.NewInt(0).Mul(x, y)
}
答案1
得分: 19
BigInt包允许您以对数时间计算x^y(由于某种原因它被称为exp)。您只需要将nil
作为最后一个参数传递即可。
package main
import (
"fmt"
"math/big"
)
func main() {
fmt.Println(new(big.Int).Exp(big.NewInt(5), big.NewInt(20), nil))
}
如果您想自己计算它,请看一下我的实现:
func powBig(a, n int) *big.Int{
tmp := big.NewInt(int64(a))
res := big.NewInt(1)
for n > 0 {
temp := new(big.Int)
if n % 2 == 1 {
temp.Mul(res, tmp)
res = temp
}
temp = new(big.Int)
temp.Mul(tmp, tmp)
tmp = temp
n /= 2
}
return res
}
或者在Go Playground上尝试一下。
英文:
BigInt package allows you to calculate x^y in log time (for some reason it is called exp). All you need is to pass nil
as a last parameter.
package main
import (
"fmt"
"math/big"
)
func main() {
fmt.Println(new(big.Int).Exp(big.NewInt(5), big.NewInt(20), nil))
}
If you are interested how to calculate it by yourself, take a look at my implementation:
func powBig(a, n int) *big.Int{
tmp := big.NewInt(int64(a))
res := big.NewInt(1)
for n > 0 {
temp := new(big.Int)
if n % 2 == 1 {
temp.Mul(res, tmp)
res = temp
}
temp = new(big.Int)
temp.Mul(tmp, tmp)
tmp = temp
n /= 2
}
return res
}
or play with it on go playground.
答案2
得分: 13
例如,
package main
import (
"fmt"
"math/big"
)
func main() {
z := new(big.Int).Exp(big.NewInt(2), big.NewInt(100), nil)
fmt.Println(z)
}
输出:
1267650600228229401496703205376
由于它是2的幂,你也可以进行位移操作:
package main
import (
"fmt"
"math/big"
)
func main() {
z := new(big.Int).Lsh(big.NewInt(1), 100)
fmt.Println(z)
}
输出:
1267650600228229401496703205376
英文:
For example,
package main
import (
"fmt"
"math/big"
)
func main() {
z := new(big.Int).Exp(big.NewInt(2), big.NewInt(100), nil)
fmt.Println(z)
}
Output:
1267650600228229401496703205376
Since it's a power of two, you could also do a bit shift:
package main
import (
"fmt"
"math/big"
)
func main() {
z := new(big.Int).Lsh(big.NewInt(1), 100)
fmt.Println(z)
}
Output:
1267650600228229401496703205376
答案3
得分: 2
如果 power % 2 == 0
,你立即返回。相反,你只想得到 base ** (power / 2)
的 result
。然后将 result * result
相乘,如果 power
是偶数,则将 base
乘以该结果。
英文:
You are returning immediately if power % 2 == 0
. Instead, you just want to get the result
of base ** (power /2)
. Then multiply result * result
, and if power
is even then multiply base
to that.
答案4
得分: 2
计算2的100次方
package main
import (
"fmt"
"math/big"
)
func main() {
n := big.NewInt(0)
fmt.Println(n.SetBit(n, 100, 1))
}
英文:
To compute 2^100
package main
import (
"fmt"
"math/big"
)
func main() {
n := big.NewInt(0)
fmt.Println(n.SetBit(n, 100, 1))
}
答案5
得分: 1
package main
import (
"fmt"
"math/big"
)
func main() {
bigx, power10 := new(big.Int), new(big.Int)
var x int64
bigx.SetInt64(x) //将int64类型的x设置给bigx
power10.Exp(big.NewInt(10), bigx, nil) //power10 *big.Int指向解决方案
str10 := power10.Text(10)
fmt.Printf(str10) //打印出数字并自行检查
}
package main
import (
"fmt"
"math/big"
)
func main() {
bigx, power10 := new(big.Int), new(big.Int)
var x int64
bigx.SetInt64(x) //将int64类型的x设置给bigx
power10.Exp(big.NewInt(10), bigx, nil) //power10 *big.Int指向解决方案
str10 := power10.Text(10)
fmt.Printf(str10) //打印出数字并自行检查
}
英文:
package main
import(
"fmt"
"math/big"
)
func main() {
bigx, power10 := new(big.Int), new(big.Int)
var x int64
bigx.SetInt64(x) //set x int64 to bigx
power10.Exp(big.NewInt(10), bigx, nil) //power10 *big.Int points to solution
str10 := power10.Text(10)
fmt.Printf(str10) // print out the number and check for your self
}
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