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Cottontail Climb (Part 1)暴力枚举Cottontail Climb (Part 2)暴力枚举Problem B: Four in a Burrow记忆化搜索Problem C: Bunny Hopscotch统计todo

3209: 花神的数论题Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 33  Solved: 18[Submit][Status][Discuss]Description背景众所周知，花神多年来凭借无边的神力狂虐各大 OJ、OI、CF、TC …… 当然也包括 CH 啦。描述话说花神这天又来讲课了。课后照例有超级难

1057. Amount of DegreesTime limit: 1.0 secondMemory limit: 64 MBCreate a code to determine the amount of integers, lying in the set [X;Y] and being a sum of exactlyK different integer de

1026: [SCOI2009]windy数Time Limit: 1 Sec  Memory Limit: 162 MBSubmit: 1280  Solved: 558[Submit][Status][Discuss]Descriptionwindy定义了一种windy数。 不含前导零且相邻两个数字之差至少为2的正整数被称为windy数。 windy想知道，在A和B之间

IGAME - Interesting Game no tags Alice and Bob play an interesting game and the game is played on a number.So a player, on his chance, can choose any non zero digit of the number and decrease the di

A. 阿里云秘钥池反演+数位dp#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> #inc

令fi,j=(i在j进制下回文)?j:1f_{i,j}=(i在j进制下回文)?j:1 求∑Ri=L∑rj=lf(i,j)\sum_{i=L}^{R}\sum_{j=l}^{r}f(i,j) 1<=L,R<=1e18,2<=j<=361<=L,R<=1e18,2<=j<=36 数位dp， 由于j进制的长度为l的回文串长度能直接算出来(j−1)j⌊l+12⌋(j-1)j^{\lfloor \f

f(n,k)=k进制下n的数位和+位数-2分别考虑几种情况并check#include&lt;bits/stdc++.h&gt; using namespace std;#define For(i,n) for(int i=1;i&lt;=n;i++)#define Fork(i,k,n) for(int i=k;i&lt;=n;i++)#define ForkD(i,k,n) fo...

A.迷宫有一个 nnn 个点 n−1n-1n−1 条边的无向连通图迷宫，其中有些点上面有人.现在所有人的目标都是逃离这个迷宫，而迷宫的出口是 1 号点，每一时刻，会依次发生以下的事情：1.在点 x 上的人选择一个点 f(x)f(x)f(x) 作为目标，要求 f(x)f(x)f(x) 必须是 xxx，或者与 xxx 有边相连的点，且对于 x≠yx\neq yx̸​=y,有 f(x)≠f(y)f...