/*
ID: tausiq11
PROG: beads
LANG: C++
*/
#include <stdio.h>
#include <iostream>
#include <list>
using namespace std;
int length;
list <char> l;
int call ()
{
int count = 0;
char temp = 'w';
bool flag = true;
do {
if ( flag && l.front () != 'w' ) {
temp = l.front ();
flag = false;
}
l.pop_front ();
count++;
//cout << l.front ();
} while ( l.size () && ( l.front () == 'w' || l.front () == temp ));
temp == 'r' ? temp = 'b' : temp = 'r' ;
while ( l.size () && ( l.back () == 'w' || l.back () == temp ) ) {
l.pop_back ();
count++;
}
return count;
}
int main ()
{
freopen ("beads.in", "r", stdin);
freopen ("beads.out", "w", stdout);
scanf ("%d", &length);
char a Error: [360] shortcode requires URL to be set;
scanf ("%s", a);
int i, j, k;
int max = 0;
int value;
for ( i = 0; i < length; i++ ) {
for ( j = i; j < length; j++ )
l.push_back ( a [j] );
for ( k = 0; k < i; k++ )
l.push_back ( a [k] );
value = call ();
if ( max < value )
max = value;
l.clear ();
}
printf ("%d\n", max);
return 0;
}

### Like this:

Like Loading...

*Related*

Sir,

Could u explain me how you had solved the problem..?

hi,

suppose u have a necklace of N beads

and u have to cut it in a certain point thus u can get maximum number of beads .. right?

N is at most as 350

so u can try out every possible points

at first i made a list/string of beads by break the necklace at point 0

and then function call () tells me how much beads i can collect from this list / string

and thus i can determine the highest number of beads

for example, given

wrbbrb

first try out, wrbbrb

then try, rbbrbw

then try, bbrbwr

then try, brbwrb

then try, rbwrbb

then try, bwrbbr

among those possible cuts u will find the highest number of beads

invoke the function call for every possible cuts

and function call will return you the number of beads u can collect for that particular cuts

i hope if you think a bit then you will find a clue about the implementation of function call

u don’t have to do exactly as i did ..

let me know if you face any problem .. good luck🙂

thnx for this explanation