Question

我需要生成一些大整数。见下面的例子。

Input   Result    
 40 165580141    
 80 37889062373143906    
120 8670007398507948658051921    
160 1983924214061919432247806074196061    
200 453973694165307953197296969697410619233826

这是我的Fortran代码：

program cycle
    use iso_fortran_env
  implicit none
  character(200) :: str
  integer :: n
    integer(kind=int64) :: x1, result, x2, x3

    do n = 40, 500, 40
        x1 = n
        result = 1
        x2 = 0
        x3 = 1
        do 
            if (x1 > 1) then
                x2 = result
                result = result + x3 
                x3 = x2     
                x1 = x1 - 1
            else
                exit
            end if
        end do
        write(str,'(i64)')  result
        print *, n, adjustl(str)
    end do
end program cycle

以下是示例输出：

 40 165580141 
 80 37889062373143906 
120 790376311979428689  
160 9217463444206948445 
200 3721511182311577122

正如您所看到的，它获得了前两个数字，但其余数字超出了64位整数的范围。我已经查看了其他问题（1），但我对一种简单的方式感兴趣，最好是在语言本身内置。在Ruby和Go中我遇到了一些麻烦。我在Fortran中忽略了一些明显的东西吗？我可以在代码中使用更好的选项吗？

Answer 1

没有内置的“大数字”支持，但我们可以先检查是否有更大的整数类型（如上面的francescalus和许多以前的页面（例如this page）所述。）使用gfortran-6.1的计算机，编译器似乎支持128位整数类型，所以我可以计算结果直到n = 160左右。

program cycle
...
integer, parameter :: verylong = selected_int_kind(32)
integer(verylong) :: x1, result, x2, x3

print *, "int32 = ", int32   !! from iso_fortran_env
print *, "int64 = ", int64
print *
print *, "kind..(16) => ", selected_int_kind(16)  !! 8
print *, "kind..(32) => ", selected_int_kind(32)  !! 16
print *, "kind..(40) => ", selected_int_kind(40)  !! -1 (not available)
print *, "kind..(64) => ", selected_int_kind(64)  !! -1 (not available)
print *
print *, "sizeof(x1)       = ", sizeof(x1), "(bytes)"       !! GNU extension
print *, "storage_size(x1) = ", storage_size(x1), "(bits)"  !! F2008
print *, "huge(x1)         = ", huge(x1)                    !! largest integer
...

结果：

 int32 =            4
 int64 =            8

 kind..(16) =>            8
 kind..(32) =>           16
 kind..(40) =>           -1
 kind..(64) =>           -1

 sizeof(x1)       =                    16 (bytes)
 storage_size(x1) =          128 (bits)
 huge(x1)         =  170141183460469231731687303715884105727

 n=          40 res= 165580141
 n=          80 res= 37889062373143906
 n=         120 res= 8670007398507948658051921
 n=         160 res= 1983924214061919432247806074196061
 n=         200 res= 37016692776042937155243383431825151522
 n=         240 res= -159769225356713774587328406036589956191
 ...

虽然没有内置的“BigInt”类型，但使用外部库（例如，从fmlib链接的this page）相当简单。由于各种运算符和赋值都过载，因此几乎不需要对代码进行任何修改。

程序：

1）下载FM.95，FMZM90.f95和FMSAVE.f95（应通过删除“.txt”来更改文件名。）

2）将库文件设为

gfortran -c -O2 FMSAVE.f95 FMZM90.f95 FM.f95
ar rv fmlib.a FM*.o

3）修改您的代码如下（修改后的部分用箭头标记）。

program cycle
    use FMZM           !<----- a module for handling big numbers
    implicit none
    character(200) :: str
    integer :: n
    type(IM) :: x1, result, x2, x3     !<----- IM = BigInt, FM = BigFloat

    do n = 40, 500, 40
        x1 = n
        result = 1
        x2 = 0
        x3 = 1
        do 
            if (x1 > 1) then
                x2 = result
                result = result + x3 
                x3 = x2     
                x1 = x1 - 1
            else
                exit
            end if
        end do
        str = IM_format( 'i200', result )   !<----- convert BigInt to string
        print *, n, trim( adjustl(str) )    !<----- print huge integers
    end do
end program cycle

4）编译（假设“test.f90”是上面的代码）：

gfortran test.f90 fmlib.a
./a.out

5）结果

   n result
  40 165580141
  80 37889062373143906
 120 8670007398507948658051921
 160 1983924214061919432247806074196061
 200 453973694165307953197296969697410619233826
 240 103881042195729914708510518382775401680142036775841
 280 23770696554372451866815101694984845480039225387896643963981
 320 5439356428629292972296177350244602806380313370817060034433662955746
 360 1244666864935793005828156005589143096022236302705537193166716344690085611761
 400 284812298108489611757988937681460995615380088782304890986477195645969271404032323901
 440 65172495098135102433647404982700073500075401759827878315356483347951218369680224170989749666
 480 14913169640232740127827512057302148063648650711209401966150219926546779697987984279570098768737999681

我们可以通过注意result n实际上等于fibonacci(n+1)来验证结果，例如我们n = 480 fibonacci(481)。< / p>

如何在Fortran中执行大整数？

1 个答案: