Polycarp likes to play with numbers. He takes some integer number xx, writes it down on the board, and then performs with it n−1n−1operations of the two kinds: divide the number xx by 33 (xx must be divisible by 33); multiply the number xx by 22. A…

传送门 D. Divide by three, multiply by two •题意 给你一个数 x,有以下两种操作,x 可以任选其中一种操作得到数 y 1.如果x可以被3整除,y=x/3 2.y=x*2 y 再执行上述两种操作的一种得到数 z: 接着对 z 得到...... 这样依次执行了 n-1 次会得到 n 个数: 现在给你这 n 个数,让你按照上述规则给这 n 个数排序,使得其满足 a1=x , a2=y , a3=z , ........ •思路 对于任意一个数 p,能通过两类操作得…

题目链接:codeforces 792C. Divide by Three 今天队友翻了个大神的代码来问,我又想了遍这题,感觉很好,这代码除了有点长,思路还是清晰易懂,我就加点注释存一下...分类吧.删除一个数字模3为M的或删除两个模3为3-M的(还有一些要删零),具体看代码. #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<strin…

Polycarp likes to play with numbers. He takes some integer number xx, writes it down on the board, and then performs with it n−1n−1 operations of the two kinds: divide the number xx by 33 (xx must be divisible by 33); multiply the number xx by 22. Af…

题意 有nnn个无序的数,对这些数进行排列,要求ai=3×ai+1a_i=3\times a_{i+1}ai​=3×ai+1​或2×ai=ai+12\times a_i=a_{i+1}2×ai​=ai+1​,保证这些数有解,输出排序后的数 AC代码 /************************************************************************* > File Name: DD.cpp > Author: WZY > QQ: 269709…

题意:给你一个长度为\(n\)的序列\(a\).对它重新排列,使得\(a_{i+1}=a_{i}/3\)或\(a_{i+1}=2*a_{i}\).输出重新排列后的序列. 题解:经典DFS,遍历这个序列,对每个数dfs,每次dfs使\(len+1\),当\(len=n\)时,说明这个序列已经满足条件了,输出并且标记一下,防止还有另外的情况导致多输出. 代码: #include <iostream> #include <cstdio> #include <cstring>…

Polycarp likes to play with numbers. He takes some integer number x, writes it down on the board, and then performs with it n−1 operations of the two kinds: 1.divide the number x by 3 (x must be divisible by 3);    2.multiply the number x by 2. After…

C. Divide by Three time limit per test: 1 second memory limit per test: 256 megabytes input: standard input output: standard output A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform…

题目链接:http://codeforces.com/problemset/problem/1176/A 思路:贪心,对第二个操作进行俩次等于将n变成n/3,第三个操作同理,我们将n不断除以2,再除以3,最后除以5,判断最后是否等于1即可. AC代码: #include<iostream> using namespace std; long long n,ans; int main(){ int T; cin >> T; while(T--){ cin >> n; an…

Codeforces 题目传送门 & 洛谷题目传送门 显然,直接暴力枚举是不可能的. 考虑将点按横纵坐标奇偶性分组,记 \(S_{i,j}=\{t|x_t\equiv i\pmod{2},y_t\equiv j\pmod{2}\}(i,j\in[0,1])\),说白了就是横坐标为偶数.纵坐标为偶数:横坐标为偶数.纵坐标为奇数:横坐标为奇数.纵坐标为偶数:横坐标为奇数.纵坐标为奇数的点集. 然后考虑以下算法: 若 \(S_{0,0},S_{1,1}\) 以及 \(S_{0,1},S_{1,0}\)…