how to change this code to be able to solve this challenge? - c

Question

I'm new in programming and I'm trying to solve some problems from online websites. And one of this problems stopped me.
You can download problem here : https://uva.onlinejudge.org/external/101/10191.pdf
I don't know how to change this code in order to determine itself where to end reading data from console.
here is the code that I wrote :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_LINE 256
int second_time(char a[])
{
int hour1, minute1, time;
hour1 = a[6] - '0';
hour1 *= 10;
hour1 += a[7] - '0';
minute1 = a[9] - '0';
minute1 *= 10;
minute1 += a[10] - '0';
time = (60*hour1) + minute1;
return time;
}
int first_time(char b[])
{
int hour2, minute2, time;
hour2 = b[0] - '0';
hour2 *= 10;
hour2 += b[1] - '0';
minute2 = b[3] - '0';
minute2 *= 10;
minute2 += b[4] - '0';
time = (60*hour2) + minute2;
return time;
}
int main(void) {
char line[MAX_LINE];
int i, j, s, k;
int first, second, delta;
scanf("%i", &s);
getchar();
int max[4], max_hour[4];
int flag = 1;
for(k=0; k<4; k++)
{
max[k] = 0;
}
while(flag)
{
for(i=0;i<s-1; i++)
{
if(i%2 == 0)
{
scanf ("%[^\n]%*c", line);
second = second_time(line);
if(s%2 == 0)
{
i++;
}
}
scanf ("%[^\n]%*c", line);
first = first_time(line);
delta = first - second;
if (delta > max[k])
max_hour[k] = second;
max[k] = delta;
second = second_time(line);
}
}
for(k=0; k<4; k++)
{
if(delta < 60)
printf("Day #%i: the longest nap starts at %.2i:%.2i and will last for %i minutes.\n", k+1, max_hour[k]/60, max_hour[k]%60, max[k]);
else
printf("Day #%i: the longest nap starts at %.2i:%.2i and will last for %i hours and %i minutes.\n", k+1, max_hour[k]/60, max_hour[k]%60, max[k]/60, max[k]%60);
}
return 0;
}

Answer 1

I've got AC with the following code
Code:
#include <stdio.h>
int main()
{
int s, c, day = 0, z, cp, cb, m;
int f[480], beg, max;
int h1, m1, h2, m2;
int p1, p2;
char line[256];
while (scanf("%d\n", &s) != EOF) {
++day;
for (z = 0; z < 480; ++z) f[z] = 1;
beg = 0;
max = 0;
for (c = 0; c < s; ++c) {
scanf("%d:%d %d:%d", &h1, &m1, &h2, &m2);
fgets(line, 256, stdin);
p1 = (h1 - 10) * 60 + m1;
p2 = (h2 - 10) * 60 + m2;
for (cp = p1; cp < p2; ++cp) f[cp] = 0;
}
c = 0;
while (c < 480) {
if (f[c]) {
cb = c;
m = 0;
while (f[c] && c < 480) {
++c;
++m;
}
if (m > max) {
beg = cb;
max = m;
}
} else {
++c;
}
}
printf("Day #%d: the longest nap starts at %02d:%02d and will last for ", day, 10 + beg / 60, beg % 60);
if (max > 59) printf("%d hours and ", max / 60);
printf("%d minutes.\n", max % 60);
}
return 0;
}

Answer 2

The function scanf() returns the number of items successfully matched. Use this return value in order to set your infinite loop and check whether you've reached EOF. Moreover, for these problems, you should usually print the results instance by instance, you do not need to keep all the answers in a vector (you don't know how large it will be anyway).

Answer 3

You have assumed the schedules are given in sorted order. It is not. So you need to sort it and then just linearly check the maximum free gap.
When to stop reading data?
Just takes input like this
while(scanf("%d",&s)!=EOF)
{
read s schedule.
sort
get maximum free gap
****BINGO****
....
}
Note: sorting is not relevant to looping but to your AC solution it is.
Code
struct schedule
{
int start;
int end;
}
struct schedule ss[MAX+1];
while(scanf("%d",&s)!=EOF)----> This is when you know you have to stop when it return EOF
{
First element of ss I would set as ss[0]={6000,6000}
last element of ss I would set as ss[s]={1080,1080}
Get s schedules and put them in ss
sort them
for each of the elements starting from 1 to s
get max ( ss[i].start - ss[i-1].end)
and corresponding end time.
get maximum free gap
****BINGO****
....
}
As you insisted your code
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_LINE 256
int second_time(char a[])
{
int hour1, minute1, time;
hour1 = a[6] - '0';
hour1 *= 10;
hour1 += a[7] - '0';
minute1 = a[9] - '0';
minute1 *= 10;
minute1 += a[10] - '0';
time = (60*hour1) + minute1;
return time;
}
int first_time(char b[])
{
int hour2, minute2, time;
hour2 = b[0] - '0';
hour2 *= 10;
hour2 += b[1] - '0';
minute2 = b[3] - '0';
minute2 *= 10;
minute2 += b[4] - '0';
time = (60*hour2) + minute2;
return time;
}
int main(void) {
char line[MAX_LINE];
int i, j, s, k;
int first, second, delta;
int max[200], max_hour[200];
int flag = 1;
for(k=0; k<200; k++)
{
max[k] = 0;
}
while(scanf("%i", &s)!=EOF)
{
getchar();
for(i=0;i<s-1; i++)
{
if(i%2 == 0)
{
scanf ("%[^\n]%*c", line);
second = second_time(line);
if(s%2 == 0)
{
i++;
}
}
scanf ("%[^\n]%*c", line);
first = first_time(line);
delta = first - second;
if (delta > max[k])
max_hour[k] = second;
max[k] = delta;
second = second_time(line);
}
for(k=0; k<s; k++)
{
if(delta < 60)
printf("Day #%i: the longest nap starts at %.2i:%.2i and will last for %i minutes.\n", k+1, max_hour[k]/60, max_hour[k]%60, max[k]);
else
printf("Day #%i: the longest nap starts at %.2i:%.2i and will last for %i hours and %i minutes.\n", k+1, max_hour[k]/60, max_hour[k]%60, max[k]/60, max[k]%60);
}
}
return 0;
}
Correct solution
typedef struct {
int st, ed;
} schedule;
int cmp(const void *i, const void *j) {
schedule *a, *b;
a = (schedule *)i, b = (schedule *)j;
return a->st <= b->st;
}
int main() {
int n, i, a, b, c, d, day = 0;
schedule TT[100];
while(scanf("%d", &n) !=EOF) {
for(i = 0; i < n; i++) {
scanf("%d:%d %d:%d", &a, &b, &c, &d);
getchar();
TT[i].st = a*60 + b;
TT[i].ed = c*60 + d;
}
qsort(TT, n, sizeof(schedule), cmp);
int beg = 600, ans = 0, point;
for(i = 0; i < n; i++) {
if(( TT[i].st-beg ) > ans)
ans = (TT[i].st-beg ), point = beg;
beg = TT[i].ed;
}
if((1080 - beg) > ans)
ans = (1080- beg), point = beg;
printf("Day #%d: the longest nap starts at ", ++day);
printf("%02d:%02d and will last for ", point/60, point%60);
if(ans >= 60)
printf("%d hours and ", ans/60);
printf("%d minutes.\n", ans%60);
}
return 0;
}