[原创代码] 无忧公主项数定律

#=
julia解欧拉计划025 1000-digit Fibonacci number.jl
2016年4月9日 15:33:05 codegay
两个方法解法一样，一个是比较字符数量，一个是比较数字大小。
=#

counter=1
a=big(0)
b=big(1)

while length(string(b))<1000
    a,b=b,a+b
    counter=counter+1
end
println(counter)
#=
4782
[Finished in 3.7s]
=#

function ff2()
    f1=big(0)
    f2=big(1)
    counter=1
    while f2<big(10)^999
        f1,f2=f2,f1+f2
        counter+=1
    end
    println(counter)
end

@time ff2()
#=
4782
  0.050925 seconds (90.95 k allocations: 4.712 MB, 54.44% gc time)
[Finished in 4.4s]
=#
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REM 解无忧公主的编程挑战004：第一个1000位的斐波那契数的项数是
@echo off&setlocal enabledelayedexpansion
set "Fa[1]=00000001"
set "Fb[1]=00000001"
for /l %%i in (2 1 125) do (
	set "Fa[%%i]=00000000"
	set "Fb[%%i]=00000000"
)
for /l %%n in (3 1 5000) do (
	set add=0
	for /l %%i in (1 1 125) do (
		set/a Fn[%%i]=1!Fa[%%i]!+1!Fb[%%i]!
		set/a tmp=Fn[%%i]+add
		set Fa[%%i]=!Fb[%%i]!
		set Fb[%%i]=!tmp:~1!
		if !tmp! geq 300000000 (set add=1) else (set add=0)
	)
	if not "!Fb[125]:~0,1!"=="0" (set/p =%%n&exit)
)
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@echo off&setlocal enabledelayedexpansion
set /a len=9,n[0]=1,n[1]=2
for /l %%n in (4 1 5000) do (
	set /a n[0]=n[1],n[1]+=!n[0]!
	if !n[1]! gtr 999999999 (
		set /a len+=1,n[0]/=10,n[1]/=10
		if !len!==1000 (set/p =%%n&exit)
	)
)
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@echo off&setlocal enabledelayedexpansion
set /a len=9,n[0]=1,n[1]=2
(for /l %%n in (4 1 5000) do (
	set /a "n[0]=n[1],n[1]+=!n[0]!,1/^!((n[1]=last+n[0])/1000000000)"||(
		set /a "1/(1000-(len+=1)),n[0]/=10,n[1]/=10"||set/p =%%n&&exit
	)
))2>nul
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@echo off&setlocal enabledelayedexpansion
set /a len=9,n[0]=1,n[1]=2
(for /l %%n in (4 1 5000) do (
	set /a "n[0]=n[1],n[1]+=!n[0]!,1/((n[1]=last+n[0])/1000000000),len+=1,n[0]/=10,n[1]/=10"&&(
		if !len!==1000 set/p =%%n&&exit
	)
))2>nul
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