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我需要生成一些大整数。请参阅下面的示例。
Input Result
40 165580141
80 37889062373143906
120 8670007398507948658051921
160 1983924214061919432247806074196061
200 453973694165307953197296969697410619233826
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
40 165580141
80 37889062373143906
120 790376311979428689
160 9217463444206948445
200 3721511182311577122
最佳答案
没有内置的“大数”支持,但我们可以先检查是否有更大的整数类型可用(如 Francescalus 上面提到的以及许多以前的页面(例如 this page )。在我的电脑上使用 gfortran-6.1 ,编译器似乎支持 128 位整数类型,所以我可以计算结果高达 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
...
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
...
FM.95
,
FMZM90.f95
, 和
FMSAVE.f95
gfortran -c -O2 FMSAVE.f95 FMZM90.f95 FM.f95
ar rv fmlib.a FM*.o
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
gfortran test.f90 fmlib.a
./a.out
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
result
来验证结果为
n
实际上等于
fibonacci(n+1) ,例如我们有
fibonacci(481)为
n = 480
.
关于fortran - 如何在 Fortran 中处理大整数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58280723/
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