代码搜索：implicit
找到约 5,250 项符合「implicit」的源代码
代码结果 5,250
www.eeworm.com/read/258562/11854236
f90 ex0608.f90
program ex0608
implicit none
integer :: i,j
outter: do i=1,3
inner: do j=1,3
write(*, "('(',i2,',',i2,')')" ) i,j
end do inner
end do outter
stop
end
www.eeworm.com/read/258562/11854238
f90 ex0606.f90
program ex0606
implicit none
integer :: dest = 9
integer floor
do floor=1, dest
if ( floor==4 ) cycle
write(*,*) floor
end do
stop
end
www.eeworm.com/read/258562/11854245
f90 ex0603.f90
program ex0603
implicit none
integer i,j
do i=1, 3
do j=1, 3
write(*, "(I2,I2)") i,j
end do
write(*,*) "another circle"
end do
stop
end
www.eeworm.com/read/258562/11854248
f90 ex0602.f90
program ex0602
implicit none
integer, parameter :: limit=10 ! 计数器的上限
integer counter ! 计数器
integer :: ans = 0 ! 拿来累加使用
do counter=2, limit ,2
ans = ans + counter
end do
www.eeworm.com/read/258562/11854255
for ex0602.for
PROGRAM ex0602
IMPLICIT NONE
INTEGER limit
PARAMETER(limit=10)
INTEGER counter
INTEGER ans
DATA ans /0/
DO 100, counter=2, limit, 2
100 ans = ans + counter
WRITE(*,*) ans
www.eeworm.com/read/258562/11854258
f90 ex0610.f90
program ex0610
implicit none
integer counter
integer :: ans = 0
do counter = 1, 100
ans = ans + counter
end do
write(*,*) ans
stop
end
www.eeworm.com/read/258562/11854265
f90 ex0611.f90
program ex0611
implicit none
integer counter
integer :: fn_1 = 1
integer :: fn_2 = 0
integer :: fn = 0
write(*,*) fn_2
write(*,*) fn_1
do counter = 3, 10
fn = fn_2 + fn
www.eeworm.com/read/258562/11854275
f90 ex0604.f90
program ex0604
implicit none
integer, parameter :: limit=10 ! 计数器的上限
integer counter ! 计数器
integer :: ans = 0 ! 拿来累加使用
counter = 2 ! 设定计数器初值
do while( counter
www.eeworm.com/read/258562/11854278
f90 zreal.f90
program main
use IMSL
implicit none
integer, parameter :: ITMAX = 100
integer, parameter :: NROOT = 2
real, parameter :: EPS = 1.0E-6
real, parameter :: ERRABS = 1.0E-6
real, para
www.eeworm.com/read/258562/11854292
f90 bvpfd1.f90
! IMSL BVPFD示范
! 求解y'''-y''+y'-y=0
! 经过y'=y, y2=y', y3=y''代换后会变成First order ODE
! y1'=y2
! y2'=y3
! y3'=y3-y2+y1
! 答案为y1=cos(t), y2=-sin(t), y3=-cos(t)
program main
use IMSL
implicit none