1.
Translate the following C code to a Fortran code:
#include <stdio.h>
#include <math.h>
double f(double t, double y)
{return y;}
main()
{
double h=0.1, t, y, k1,k2,k3,k4;
int i;
/* initial value */
t=0.0; y=1.0;
for (i=0; i<=10; i++)
{
printf("t= %lf rk= %lf exact=%lf\n", t, y, exp(t));
k1=h*f(t,y);
k2=h*f(t+h/2, y+k1/2.0);
k3=h*f(t+h/2, y+k2/2.0);
k4=h*f(t+h, y+k3);
y= y+(k1+2.0*k2+2.0*k3+k4)/6.0;
t=t+h;
}
return 0;
} |
You have show (1) your code and (2) an example run.
2.
Execute the following program and figure out what it does.
#include <stdio.h>
int main(t,_,a)
char *a;
{return!0<t?t<3?main(-79,-13,a+main(-87,1-_,
main(-86, 0, a+1 )+a)):1,t<_?main(t+1, _, a ):3,main ( -94, -27+t, a
)&&t == 2 ?_<13 ?main ( 2, _+1, "%s %d %d\n" ):9:16:t<0?t<-72?main(_,
t,"@n'+,#'/*{}w+/w#cdnr/+,{}r/*de}+,/*{*+,/w{%+,/w#q#n+,/#{l,+,/n{n+\
,/+#n+,/#;#q#n+,/+k#;*+,/'r :'d*'3,}{w+K w'K:'+}e#';dq#'l q#'+d'K#!/\
+k#;q#'r}eKK#}w'r}eKK{nl]'/#;#q#n'){)#}w'){){nl]'/+#n';d}rw' i;# ){n\
l]!/n{n#'; r{#w'r nc{nl]'/#{l,+'K {rw' iK{;[{nl]'/w#q#\
n'wk nw' iwk{KK{nl]!/w{%'l##w#' i; :{nl]'/*{q#'ld;r'}{nlwb!/*de}'c \
;;{nl'-{}rw]'/+,}##'*}#nc,',#nw]'/+kd'+e}+;\
#'rdq#w! nr'/ ') }+}{rl#'{n' ')# }'+}##(!!/")
:t<-50?_==*a ?putchar(a[31]):main(-65,_,a+1):main((*a == '/')+t,_,a\
+1 ):0<t?main ( 2, 2 , "%s"):*a=='/'||main(0,main(-61,*a, "!ek;dc \
i@bK'(q)-[w]*%n+r3#l,{}:\nuwloca-O;m .vpbks,fxntdCeghiry"),a+1);
}
