HDUOJ-----1066Last non-zero Digit in N!】的更多相关文章

Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... Note: n is positive and will fit within the range of a 32-bit signed integer (n < 231). Example 1: Input: 3 Output: 3 Example 2: Input: 11 Output: 0 Explanatio…
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example: Given n = 13, Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13. Hint: Beware of overflow.…
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example:Given n = 13,Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13. Hint: Beware of overflow. 大…
C. Digit Tree time limit per test:3 seconds memory limit per test:256 megabytes input:standard input output:standard output ZS the Coder has a large tree. It can be represented as an undirected connected graph of n vertices numbered from 0 to n - 1 a…
date:2016-09-13 今天开始注册了kaggle,从digit recognizer开始学习, 由于是第一个案例对于整个流程目前我还不够了解,首先了解大神是怎么运行怎么构思,然后模仿.这样的学习流程可能更加有效,目前看到排名靠前的是用TensorFlow.ps:TensorFlow是可以直接安linux环境下面,但是目前不能在windows环境里面运行(伤心一万点). TensorFlow模块用的是NN(神经网络),既然现在接触到可以用神经网络的例子我再也不好意思再逃避学习神经网络下面…
http://acm.hdu.edu.cn/showproblem.php?pid=1455 http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=243 uva开头描述: 307 - Sticks Time limit: 3.000 seconds hduoj 开头描述: E - Sticks Time Limit:3000MS     Memo…
Problem Introduction The Fibonacci numbers are defined as follows: \(F_0=0\), \(F_1=1\),and \(F_i=F_{i-1}+F_{i-2}\) for $ i \geq 2$. Problem Description Task.Given an integer \(n\),find the last digit of the \(n\)th Fibonacci number \(F_n\)(that is ,…
Last non-zero Digit in N! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5861    Accepted Submission(s): 1451 Problem Description The expression N!, read as "N factorial," denotes the pro…
Backward Digit Sums Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6350   Accepted: 3673 Description FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum ad…
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example:Given n = 13,Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13. Hint: Beware of overflow. i…
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example: Given n = 13, Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13. 解题思路: 递归 static public in…
前言 1.理论知识:UFLDL教程.Deep learning:十六(deep networks) 2.实验环境:win7, matlab2015b,16G内存,2T硬盘 3.实验内容:Exercise: Implement deep networks for digit classification.利用深度网络完成MNIST手写数字数据库中手写数字的识别.即:用6万个已标注数据(即:6万张28*28的图像块(patches)),作为训练数据集,然后把它输入到栈式自编码器中,它的第一层自编码器…
传送门 Leftmost Digit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15305    Accepted Submission(s): 5937 Problem Description Given a positive integer N, you should output the leftmost digit of N…
Leftmost Digit Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1982 Accepted Submission(s): 884 Problem Description Given a positive integer N, you should output the leftmost digit of N^N. Input T…
Count the number of k's between 0 and n. k can be 0 - 9. Example if n = 12, k = 1 in [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] we have FIVE 1's (1, 10, 11, 12) 分析: 利用while 循环 + num % 10可以获取 num中的所有数字. class Solution { /* * param k : As description.…
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35848    Accepted Submission(s): 13581 Problem Description Given a positive integer N, you should output the most right digit of N^N.   Input The…
Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... Note: n is positive and will fit within the range of a 32-bit signed integer (n < 231). Example 1: Input: 3 Output: 3 Example 2: Input: 11 Output: 0 Explanatio…
http://acm.hdu.edu.cn/showproblem.php?pid=4712 Hamming Distance Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 1610    Accepted Submission(s): 630 Problem Description (From wikipedia) For bina…
hduoj 4706 Children's Day 2013 ACM/ICPC Asia Regional Online —— Warmup Herding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2005    Accepted Submission(s): 563 Problem Description Little Joh…
本文以下内容来自读论文以后认为有价值的地方,论文来自:convolutional Neural Networks Applied to House Numbers Digit Classification . 对于房门号的数字识别问题,文中提出的方法是基于卷积神经网络的,卷积神经网络集特征提取与目标分类于一体,这一点有别于传统的识别方法(传统方法中一般都是基于人工设计的特征提取器,然后把提取到的特征输入给分类器). 文中在传统的卷积神经网络基础上有两点改进: 第一:pooling层,传统的方法的…
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example:Given n = 13,Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13. Hint: Beware of overflow. c…
UVa 1225 题目大意:把前n(n<=10000)个整数顺次写在一起,12345678910111213...,数一数0-9各出现多少字 解题思路:用一个cnt数组记录0-9这10个数字出现的次数,先将cnt初始化为0,接着让i从1枚举到n, 对每个i,处理以活的i的每一个位置上的数,并在相应的cnt下标上+1 最后输出cnt数组即可 /* UVa 1225 Digit Counting --- 水题 */ #include <cstdio> #include <cstring…
UVa 1583 题目大意:如果x加上x的各个数字之和得到y,那么称x是y的生成元. 给定数字n,求它的最小生成元 解题思路:可以利用打表的方法,提前计算出以i为生成元的数,设为d,并保存在a[d]中(a[d]=i),反复枚举,若是初次遇到或遇到更小的则更新 相关说明:本来按书上来,在更新数组a时,if里是有或上 i < a[y]这个条件的, 但观察到由于i是从小到大枚举的,因此只会更新一次,即第一次填进去的就是最小生成元,因此去掉仍然AC /* UVa 1583 Digit Generator…
HDU 1061 题目大意:给定数字n(1<=n<=1,000,000,000),求n^n%10的结果 解题思路:首先n可以很大,直接累积n^n再求模肯定是不可取的, 因为会超出数据范围,即使是long long也无法存储. 因此需要利用 (a*b)%c = (a%c)*(b%c)%c,一直乘下去,即 (a^n)%c = ((a%c)^n)%c; 即每次都对结果取模一次 此外,此题直接使用朴素的O(n)算法会超时,因此需要优化时间复杂度: 一是利用分治法的思想,先算出t = a^(n/2),若…
Lesson learnt: one effective solution for bit\digit counting problems: counting by digit\bit http://www.hawstein.com/posts/20.4.html class Solution { public: /* * param k : As description. * param n : As description. * return: How many k's between 0…
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example:Given n = 13,Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13. 解法:参考编程之美 132页 2.4 1的数目,以下代…
IP-Box could crack 4 digit passcode, what about 6 digit passcode??? All you need to do is to upgrade your iOS to iOS9, and you could use 6 digit passcode to make your iPhone more secure. The latest version of iOS is 9.0.1. Look at the screen and you…
这道题其实挺水,只是写的时候需要想清楚.我的方法是: 1.将[a,b]转化为[0,b+1)-[0,a) 2.预处理出非0的v在区间[0,10^p)出现次数以及0在区间[0,10^p)出现数 3.将一个区间再拆分为几段,如: 12345拆分为[0,10000),[10000,12000),[12000,12300),[12300,12340),[12340,12346) 下面是代码: #include<cstdio> using namespace std ; static class digi…
  Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11011 Accepted Submission(s): 4214 Problem Description Given a positive integer N, you should output the leftmost digit of N^N. Input The input co…
题目 Square digit chains A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before. For example, 44 → 32 → 13 → 10 → 1 → 185 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89 The…