Chapter 5 project 8 about c-solutions

[rishabhh1809]

`
#include <stdio.h>
#include <stdlib.h>

#define lowest(a,b,c,d,e,f,g,h) (a<b && a<c && a<d && a<e && a<f && a<g && a<h)
#define middletime(i,j) ((((1440-j)+i)/2)+j)
#define inmin(k,l) (60 * k + l)

int main (void)
{
int hh, mm, rr, dh, dm, d, ah, am, a,
    dhfirst, dh2, dh3, dh4, dh5, dh6, dh7, dhlast,
    dmfirst, dm2, dm3, dm4, dm5, dm6, dm7, dmlast,
    dfirst, d2, d3, d4, d5, d6, d7, dlast,
    s1, s2, s3, s4, s5, s6, s7, s8,
    ahfirst, ah2, ah3, ah4, ah5, ah6, ah7, ahlast,
    amfirst, am2, am3, am4, am5, am6, am7, amlast,
    afirst, a2, a3, a4, a5, a6, a7, alast;

printf("Enter a 24-hour time: ");
scanf("%d:%d", &hh, &mm);

dhfirst = 8;    dmfirst = 0;     dfirst = inmin(8,dmfirst);
ahfirst = 10;   amfirst = 16;    afirst = inmin(10,amfirst);

dhlast = 9;    dmlast = 45;    dlast = inmin(21,dmlast);   
ahlast = 11;   amlast = 58;    alast = inmin(23,amlast);

if (inmin(hh,mm)>dfirst && inmin(hh,mm)<dlast)
    rr = inmin(hh,mm);
else if (1440<middletime(dfirst,dlast))
    if (inmin(hh,mm)>middletime(dfirst,dlast)-1440 && inmin(hh,mm)<dfirst)
        rr = dfirst + 1;
    else
        rr = dlast - 1;
else if (1440>middletime(dfirst,dlast))
    if (inmin(hh,mm)<middletime(dfirst,dlast) && inmin(hh,mm)>dlast)
        rr = dlast - 1;
    else
        rr = dfirst + 1;
else if (1440==middletime(dfirst,dlast))
    if (inmin(hh,mm)>dlast) 
        rr = dlast - 1;
    else
        rr = dfirst + 1;
else
    rr = inmin(hh,mm);

                                                s1 = abs(dfirst-rr);
dh2 = 9;    dm2 = 43;    d2 = inmin(9,dm2);     s2 = abs(d2-rr);   ah2 = 11;   am2 = 52;    a2 = inmin(11,am2);
dh3 = 11;   dm3 = 19;    d3 = inmin(11,dm3);    s3 = abs(d3-rr);   ah3 = 1;    am3 = 31;    a3 = inmin(13,am3);
dh4 = 12;   dm4 = 47;    d4 = inmin(12,dm4);    s4 = abs(d4-rr);   ah4 = 3;    am4 = 0;     a4 = inmin(15,am4);
dh5 = 2;    dm5 = 0;     d5 = inmin(14,dm5);    s5 = abs(d5-rr);   ah5 = 4;    am5 = 8;     a5 = inmin(16,am5);
dh6 = 3;    dm6 = 45;    d6 = inmin(15,dm6);    s6 = abs(d6-rr);   ah6 = 5;    am6 = 55;    a6 = inmin(17,am6);
dh7 = 7;    dm7 = 0;     d7 = inmin(19,dm7);    s7 = abs(d7-rr);   ah7 = 9;    am7 = 20;    a7 = inmin(21,am7);
                                                s8 = abs(dlast-rr);

if (lowest(s1,s2,s3,s4,s5,s6,s7,s8))
    {dh=dhfirst; dm=dmfirst; ah=ahfirst; am=amfirst; d=dfirst; a=afirst;}
else if (lowest(s2,s1,s3,s4,s5,s6,s7,s8))
    {dh=dh2; dm=dm2; ah=ah2; am=am2; d=d2; a=a2;}
else if (lowest(s3,s2,s1,s4,s5,s6,s7,s8))
    {dh=dh3; dm=dm3; ah=ah3; am=am3; d=d3; a=a3;}
else if (lowest(s4,s2,s3,s1,s5,s6,s7,s8))
    {dh=dh4; dm=dm4; ah=ah4; am=am4; d=d4; a=a4;}
else if (lowest(s5,s2,s3,s4,s1,s6,s7,s8))
    {dh=dh5; dm=dm5; ah=ah5; am=am5; d=d5; a=a5;}
else if (lowest(s6,s2,s3,s4,s5,s1,s7,s8))
    {dh=dh6; dm=dm6; ah=ah6; am=am6; d=d6; a=a6;}
else if (lowest(s7,s2,s3,s4,s5,s6,s1,s8))
    {dh=dh7; dm=dm7; ah=ah7; am=am7; d=d7; a=a7;}
else
    {dh=dhlast; dm=dmlast; ah=ahlast; am=amlast; d=dlast; a=alast;}

printf("Closest departure is ");

if (d<720 && a<720)
    printf("%d:%0.2d a.m., arriving at %d:%0.2d a.m.\n", dh, dm, ah, am);
else if (d<720 && a>720)
    printf("%d:%0.2d a.m., arriving at %d:%0.2d p.m.\n", dh, dm, ah, am);
else if (d>720 && a<720)
    printf("%d:%0.2d p.m., arriving at %d:%0.2d a.m.\n", dh, dm, ah, am);
else
    printf("%d:%0.2d p.m., arriving at %d:%0.2d p.m.\n", dh, dm, ah, am);

return 0;
}

`

[Blank0211]

Just a thought.

Not sure if yesterdays flights are to be considered, but currently a user entered time of e.g. 00:45 will give 8:00 a.m. as the closest flight even if the last flight from the day before is closer in time (3 hours back in time to 9:45 p.m. vs 7 hours 15 minutes forward to 8:00 a.m.).

if (user_time < 173) printf(...), a simple check like this in the beginning, solves the problem. i.e. if time < 2:53, which is the middle point between the two departures.

[Blank0211]

What is that code? that looks needlessly complicated. Does it even work?