the AC output is:
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S0
11
--
S0
0
--
S0
11
S0
--
00
--
S0
SSS11
SS111
SS---
SSS11
SS11
S11
-----
10101
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S0
11
--
S0
0
--
S0
11
S0
--
00
--
S0
SSS11
SS111
SS---
SSS11
SS11
S11
-----
10101
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0 11
11 0
0 0
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0
11
--
0
0
--
0
11
0
--
00
--
0
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A\n\n
A\n\n
A\n
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A\n\n
A\n\n
A\n\n <---- \n :o
this is strange. Why this happens?#include <iostream>
using namespace std;
int
main(){
cout << "something" << endl;
}
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#include <stdlib.h>
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if(even(graph) || !conected(graph))
"some beads may be lost"
else
print euler path;
this work in this problem?uzioriluzan wrote: As shahriar_manzoor intended, if there is an odd number of odd numbers, there will be no solution.
The other cases, a combination of all possibilities will do, as .. said.
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Today is: 31 12
31 12 *TODAY* New Year Eve
1 1 ******* New Year Again
1 1 ****** New Year
2 1 *** Nothing
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AC
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3
10 2
100 2
1000 2
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6
62
625