**Hint**

**Video editorial**

**Idea**

**Video editorial**

**Hint**

**Video editorial**

**Hint 1**

**Hint 2**

**Video editorial**

**Easy version**

**Hint 1**

**Hint 2**

**Video editorial**

**Hint 1**

**Hint 2**

**Hint 3**

**Video editorial**

**Hint 1**

**Hint 2**

**Video editorial**

**Daily Chat QnA**

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

1 | tourist | 4009 |

2 | jiangly | 3773 |

3 | Radewoosh | 3646 |

4 | ecnerwala | 3624 |

5 | jqdai0815 | 3620 |

5 | Benq | 3620 |

7 | orzdevinwang | 3612 |

8 | Geothermal | 3569 |

8 | cnnfls_csy | 3569 |

10 | gyh20 | 3447 |

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

1 | cry | 161 |

2 | maomao90 | 160 |

2 | awoo | 160 |

4 | atcoder_official | 157 |

5 | -is-this-fft- | 155 |

5 | nor | 155 |

7 | adamant | 153 |

8 | maroonrk | 152 |

8 | Um_nik | 152 |

10 | djm03178 | 146 |

Greedily take Cow as long as you can.

Just print all index $$$A[i*K][j*K]$$$

Bruteforce works, now analyse why it works.

Replace all 0 with -1. Now all segments with equal 0 and 1 have zero sum.

Instead of counting segments inside each range. For each substring with zero sum, count segments it is a part of.

Greedy, always take the index with maximum value of $$$A-i$$$

Binary search on the value you will add in $$$k^{th}$$$ operation.

Delete one road, now path between any pair of friends is unique. Try to count paths which do not lie between any pair of friends.

Given an array A, and multiple queries $$$L_i$$$ and $$$R_i$$$. Can you find the minimum value in this range, and how many times it appears?

Tutorial of Codeforces Round 962 (Div. 3)

↑

↓

Codeforces (c) Copyright 2010-2024 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Sep/11/2024 10:17:26 (l3).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|

The E question is almost exactly the same as the "skill upgrade" of the 13th Provincial Competition of the 2022 Blue Bridge Cup, and I have done this question before and memorized the answers in the solution, but I said that I checked the duplicate with other codes, but I don't know those people at all

Thanks for the video editorial