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  • PAT-1015 Reversible Primes2022-02-11 10:32:09

    1015 Reversible Primes (20 分) A reversible prime in any number system is a prime whose “reverse” in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime. Now given any tw

  • 1015 Reversible Primes (20 分)2021-09-01 18:34:13

    A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime. Now given any two positive integers N (&

  • boost::geometry::reversible_view用法的测试程序2021-07-12 10:03:01

    boost::geometry::reversible_view用法的测试程序 实现功能 C++实现代码 实现功能 boost::geometry::reversible_view用法的测试程序 C++实现代码 #include <algorithm> #include <iterator> #include <sstream> #

  • 用LUA刷PTA ——1015 Reversible Primes2021-06-19 14:05:45

    踩的坑 1.题目意思是十进制数转化成对应进制后翻转,得到的数是不是素数,我看了好久为啥23是二进制数。 2.测试点1,1不是素数 质数是指在大于1的自然数中,除了1和它本身以外不再有其他因数的自然数。 代码 --IO local read_func = io.read('*a'):gmatch('%p-%d+') local read_num

  • 大幅减少GPU显存占用:可逆残差网络The Reversible Residual Network2021-03-28 22:57:19

    前序:  Google AI最新出品的论文Reformer 在ICLR 2020会议上获得高分,论文中对当前暴热的Transformer做两点革新:一个是局部敏感哈希(LSH);一个是可逆残差网络代替标准残差网络。本文主要介绍变革的第二部分,可逆残差网络。先从神经网络的反向传播讲起,然后是标准残差网络,最后自然过渡

  • PTA 1015 Reversible Primes2021-03-10 15:00:48

    题目描述 分析:将一个数在k进制下翻转,判断这个数和翻转后的数是否都是素数,埃氏筛找出素数表,完后进制转换 #include<cstdio> #include<algorithm> using namespace std; const int INF=1e5; bool prime[INF]; void findPrime(){ fill(prime+2,prime+INF,true); for(

  • 1015 Reversible Primes2021-02-08 13:57:30

    1015 Reversible Primes A reversible prime in any number system is a prime whose “reverse” in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime. Now given any two positi

  • ARC111 B - Reversible Cards(思维+树的判断)2021-01-19 18:31:58

    题意: 解法: 将(ai,bi)视为ai和bi之间有一条边, 那么对于图中的所有连通块: 1.如果连通块是一棵cnt个点的树,那么一定能选出cnt-1个点. 2.如果连通块是cnt个点的非树连通图,那么一定能选出cnt个点. 因此问题变为判断每个连通块是否是树,dfs一下就行了. code: #include <bits/

  • 1015 Reversible Primes (20分)2021-01-05 09:32:05

    我丢,又读假题了。。。理解成十进制系统下的质数判断了 看到系统,不免想起被离散支配的恐惧+_+ 题意 给定两个整数 \(N\) 和 \(D\),如果 \(N\) 是一个质数,并且将 \(N\) 转化为 \(D\) 进制表示后,再进行反转,得到的新数字转化为十进制表示后如果也是一个质数,则称 \(N\) 在 \(D\) 进制系

  • 1015 Reversible Primes2020-06-24 18:02:01

    A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime. Now given any two positive integers N (&

  • 1015 Reversible Primes (20 分)2019-11-23 15:03:55

    A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime. Now given any two positive integers N (&

  • PAT甲级 1015 Reversible Primes 进制转化+反转+判断素数2019-07-15 21:03:46

    代码如下: //数的反转、判断素数、进制间的转化 #include<iostream> #include<stdio.h> using namespace std; int num[100];//记录转化为radex进制数每位的数组 bool isprime(int x){//判断素数 if(x==1){ return false; } for(int i=2;i*i<=x;i++){ if(x%

  • PAT 1015 Reversible Primes2019-05-11 16:54:52

    A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime. Now given any two positive integers N (&

  • 1015 Reversible Primes (20 分)2019-03-28 08:55:28

    1015 Reversible Primes (20 分) A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime. Now given

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