/*
ID: tausiq11
PROG: milk2
LANG: C++
*/
#include <iostream>
using namespace std;
int main () {
freopen ("milk2.in", "r", stdin);
freopen ("milk2.out", "w", stdout);
int n;
cin >> n;
int a [2] [5005];
int i;
for ( i = 0; i < n; i++ ) {
cin >> a [0] [i];
cin >> a [1] [i];
}
for ( i = 0; i < n; i++ ) {
for ( int j = i + 1; j < n; j++ ) {
if ( a [0] [i] > a [0] [j] ) {
int temp = a [0] [i];
a [0] [i] = a [0] [j];
a [0] [j] = temp;
temp = a [1] [i];
a [1] [i] = a [1] [j];
a [1] [j] = temp;
}
}
}
int start_continue = a [0] [0];
int finish_continue = a [1] [0];
int max_continue = finish_continue - start_continue;
int max_idle = 0;
for ( i = 1; i < n; i++ ) {
if ( finish_continue < a [1] [i] ) {
if ( a [0] [i] <= finish_continue )
finish_continue = a [1] [i];
else {
if ( finish_continue - start_continue > max_continue )
max_continue = finish_continue - start_continue;
if ( a [0] [i] - finish_continue > max_idle )
max_idle = a [0] [i] - finish_continue;
start_continue = a [0] [i];
finish_continue = a [1] [i];
}
}
}
printf ("%d %d\n", max_continue, max_idle);
return 0;
}

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## Published by Shahab

Completed B.Sc in CSE, @United International University, Dhaka.
Currently working as Development Engineer, Android and iOS Application.
View all posts by Shahab

In my opinion the output for input:

4

300 500

400 800

900 1100

1000 2000

should be 1100 because the longest continous time interval will be 900 to 2000, but your code is returning 500 , because its taking 300 to 800 as the longest time interval.

i think that is the right code

#include

using namespace std;

int main () {

int n;

cin >> n;

int a [2] [5005];

int i;

for ( i = 0; i > a [0] [i];

cin >> a [1] [i];

}

for ( i = 0; i < n; i++ ) {

for ( int j = i + 1; j a [0] [j] ) {

int temp = a [0] [i];

a [0] [i] = a [0] [j];

a [0] [j] = temp;

temp = a [1] [i];

a [1] [i] = a [1] [j];

a [1] [j] = temp;

}

}

}

int start_continue = a [0] [0];

int finish_continue = a [1] [0];

int max_continue = finish_continue – start_continue;

int max_idle = 0;

for ( i = 1; i < n; i++ ) {

if ( finish_continue < a [1] [i] ) {

if ( a [0] [i] max_continue )

max_continue = finish_continue – start_continue;

if ( a [0] [i] – finish_continue > max_idle )

max_idle = a [0] [i] – finish_continue;

start_continue = a [0] [i];

finish_continue = a [1] [i];

}

}

}

cout << max_continue<<" "<< max_idle;

return 0;

}