Program www_fcode_cn
Implicit None
integer :: i = 3
integer :: a(1000)!//必须定长
a(1:2)=1
Do while( i<30 .and. a(i)<100000)
a(i) = a(i-1)+a(i-2)
write(*,*) a(i)
i=i+1 !//先输出后+1
End Do
End Program www_fcode_cn
pasuka 发表于 2015-5-15 10:54
随手网上搜索就一大把参考代码
Fortran fibonacci woes - Stack Overflow
http://stackoverflow.com/questi ...
fcode 发表于 2015-5-15 20:00
修改意见参考注释:[mw_shl_code=fortran,true]Program www_fcode_cn
Implicit None
integer :: i = 3
百事可乐 发表于 2015-5-16 14:13
斐波拉契数列编程实现很简单的,还需要看文献?
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