# | User | Rating |
---|---|---|
1 | ecnerwala | 3649 |
2 | Benq | 3581 |
3 | orzdevinwang | 3570 |
4 | Geothermal | 3569 |
4 | cnnfls_csy | 3569 |
6 | tourist | 3565 |
7 | maroonrk | 3531 |
8 | Radewoosh | 3521 |
9 | Um_nik | 3482 |
10 | jiangly | 3468 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 159 |
5 | nor | 158 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 151 |
8 | SecondThread | 147 |
9 | orz | 146 |
10 | pajenegod | 145 |
pratikmoona
|
13 years ago,
#
|
0
The concept is very simple...
Assuming that there no. of boys (b) < no. of girls (g), then, we have b + 1 places where we can fills groups of girls. To minimize the number of girls in a line, we shall try to distribute them equally, i.e., g / (b + 1) girls everywhere. If g % (b + 1) != 0, then we shall distribute g % (b + 1) girls in those many places and the answer then becomes g / (b + 1) + 1. Trivial cases, g = b, answer is 1. g or b = 0, answer is b or g respectively.
→
Reply
|
Name |
---|