Sheldon Numbers 题目链接: http://acm.hust.edu.cn/vjudge/contest/127406#problem/H Description According to Sheldon Cooper, the best number is 73. In his own words, "The best number is 73. 73 is the 21st prime number. Its mirror, 37, is the 12th, and its m…

C. Hexadecimal's Numbers time limit per test 1 second memory limit per test 64 megabytes input standard input output standard output One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the me…

1565: Vampire Numbers 时间限制: 3 Sec  内存限制: 128 MB提交: 20  解决: 9[提交][状态][讨论版] 题目描述 The number 1827 is an interesting number, because 1827=21*87, and all of the same digits appear on both sides of the `='. The number136948 has the same property: 136948=14…

Problem Description Recently, Mr. Xie learn the concept of happy number. A happy number is a number contain all digit 7 or only 1 digit other than 7. For example, 777 is a happy number because 777 contail all digit 7, 7177 and 87777 both happy number…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1216 Assistance Required Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3158    Accepted Submission(s): 1662 Problem Description After the 1997/19…

UVA - 13022 Sheldon Numbers 二进制形式满足ABA,ABAB数的个数(A为一定长度的1,B为一定长度的0). 其实就是寻找在二进制中满足所有的1串具有相同的长度,所有的0串也具有相同的长度,并且在给定范围内的个数. 位运算.通过分析不难发现,所有解不会很大,因此我们可以暴力,用两个for分别枚举0串和1串的长度,然后交替放入一个值内(注意先放1),同时更新答案. 将值存入set,发现所有满足条件的个数为4809(4810). ps:因为这里的2^63大于long lon…

///找到一个数字序列包含所有n位数(连续)一次且仅一次 ///暴力打表 ///Time:141Ms Memory:2260K #include<iostream> #include<cstring> #include<cstdio> using namespace std; #define MAX 1000010 bool v[MAX]; char num[6][MAX]; int main() { //freopen("in.txt", &quo…

转载请声明出处:http://www.cnblogs.com/kevince/p/3887827.html    ——By Kevince 首先声明一下,这里的规律指的是循环,即找到最小循环周期. 这么一说大家心里肯定有数了吧,“不就是next数组性质的应用嘛”,没错,正是如此. 在ACM的比赛中有些时候会遇到一些题目,可以或必须通过找出数据的规律来编写代码,这里我们专门来讨论下 如何运用KMP中next数组的性质 来寻找一个长数组中的最小循环周期. 先来看一道题 ZOJ 3785 What d…

u Calculate e Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 46844    Accepted Submission(s): 21489 Problem Description A simple mathematical formula for e is where n is allowed to go to infini…

题意 给出一个二进制数\(n\),每次操作可以将一个整数\(x\)简化为\(x\)的二进制表示中\(1\)的个数,如果一个数简化为\(1\)所需的最小次数为\(k\),将这个数叫做特殊的数, 问从\(1\)到\(n\)一共有多少个特殊的数,答案对\(1e9+7\)取模. 分析 \(n\)最大为\(2^{1000}\),二进制表示中最多有\(1000\)个\(1\),所以\(n\)以内的数经过一次简化后将变为\(1000\)以内的数,我们可以暴力打表\(1000\)以内的数简化为\(1\)所需的最…