I need help to solved this intersting problem link .

thank you in advance.

# | User | Rating |
---|---|---|

1 | tourist | 3778 |

2 | Benq | 3592 |

3 | ecnerwala | 3521 |

4 | Um_nik | 3423 |

5 | jiangly | 3375 |

6 | Petr | 3342 |

7 | Radewoosh | 3337 |

8 | scott_wu | 3313 |

9 | maroonrk | 3265 |

10 | yosupo | 3259 |

# | User | Contrib. |
---|---|---|

1 | 1-gon | 203 |

2 | Errichto | 202 |

3 | rng_58 | 194 |

3 | SecondThread | 194 |

5 | awoo | 187 |

6 | vovuh | 183 |

7 | Um_nik | 182 |

8 | antontrygubO_o | 177 |

9 | Ashishgup | 175 |

10 | -is-this-fft- | 171 |

I need help to solved this intersting problem link .

thank you in advance.

↑

↓

Codeforces (c) Copyright 2010-2021 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Jan/16/2021 21:19:24 (f1).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|

if you now know how to solve it then please help .

not sure at all but this might work:

First, let's work on finding the # of cool numbers <= X, for a given X (then we can get the answer by computing that for B and A-1).

For a fixed X, we can get the # of cool numbers with a dp[digit][sum1][sum2][alreadyless_bit]

digit will be the position of the digit we're trying, We will go from most significative to least. sum1 and sum2 are the sums of the two sets. alreadyless_bit tells if we are already strictly less than X, if so, the new digit can be any digit, otherwise, it has to be between 0 and the respective X's digit.

This counts the cool numbers but it might count each one many times, I don't know if you can avoid that hehe

Did you solve this problem later.. ? If yes can you share your idea?

Can someone help in this question

I might be wrong, but how about...

Spoiler...submitting some "solutions" close to the source size limit? ;)

I couldn't think of any solution. I know above idea proposed by lmn0x4F seems to be good but couldn't find a way to for overcounting issue.

Can you provide some hint?

I viciously precomputed answers for blocks of size 200000 and then solved naively for the remainders. You might want to have a fast naive solver (i.e., $$$O(d)$$$ instead of $$$O(d^2)$$$ knapsack). Keep in mind the 50K source size limit.

Thanks a lot for sharing your approach