题目大意：给n个0~15之间的数，有3种更新操作，1种询问操作。3种更新操作是：1、让某个闭区间的所有数字与一个0~15之间的数字进行逻辑与运算；2、让某个闭区间的所有数字与一个0~15之间的数字进行逻辑或运算；3、让某个闭区间的所有数字与一个0~15之间的数字进行异或运算。一种询问操作是询问某个闭区间的所有数字之和。题目分析：所有的输入数字都是在0~15之间，可以二进制中的每一位建立线段树，建立四棵。3种更新操作实际上只有...
解题思路一：找规律，动态规划。f(n):长度为n的数字中包含的独立数位的数字个数。f(1) = 10 (0,1,2,3,4,...9)f(2)=9*9 ,因为十位只能是（1,2,...9），对于十位的每个数，个位只能取其余的9个数字。f(3)=f(2)*8=9*9*8f(10)=9*9*8*7...*1f(11)=0=f(12)=f(13)....class Solution { public:int countNumbersWithUniqueDigits(int n) {if(n==0)return 1;int res=10,available=9,cur=9;for(int i=1;i<n;i++){if(cur==0)break;int tmp = ...
Count the number of digits in a long integer entered by a user. int countDigit(longlong n) { int count = 0; while (n != 0) { n = n / 10; ++count; } return count; } 原文：https://www.cnblogs.com/JasperZhao/p/12814593.html
来源：http://m.blog.csdn.net/article/details?id=70861055Trung is bored with his mathematicshomeworks. He takes a piece of chalk and starts writing a sequence ofconsecutive integers starting with 1 to N (1 < N < 10000). After that, hecounts the number of times each digit (0 to 9) appears in the sequence. Forexample, with N = 13, the sequence is: 12345678910111213In this sequence, 0appears once, 1 a...
题目链接：https://leetcode.com/problems/count-numbers-with-unique-digits/题目： Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.Example: Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])Hint:A direct way is to use the backtracking approach. Backtracking should contains...
原因如题，git邮箱配置错误修改 git config --global user.name "xxx"; git config --global user.email "xxx";执行完后切记将之前已经做的add、commit操作reset 掉，然后再重新add、commit，因为add、commit 时已经记录下了做了该操作时的帐号信息。 原文：http://www.cnblogs.com/boann/p/5627082.html
[inside this square brackets give a name to the followed acc.] name = github_username email = github_emailaddress[any other name] name = github_username email = github_email[credential] helper = osxkeychain useHttpPath = true Multiple SSH keys for different accounts on Github or GitlabSSH GIT GITLAB GITHUBSometimes you need more accounts than one for access to Github or Gitlab and simi...
题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1122题意：给你m个数，选取n个数组成一个整数，使得整数各位的最大数与最小数的差小于2。问有几种选法？解法：ＤＰ。ｄｐ［ｉ］［ｊ］表示以ｊ结尾的ｉ位整数的解法数目。 答案即为ｓｕｍ（ｄｐ［ｎ］［ｋ］ （１＜＝ｋ＜＝９，且ｋ在集合Ｓ中） ）代码：#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include ...
场景$ git push To https://github.com/XXX/XXX! [rejected] dev -> dev (non-fast-forward) error: failed to push some refs to 'https://github.com/XXX/XXX' hint: Updates were rejected because the tip of your current branch is behind hint: its remote counterpart. Integrate the remote changes (e.g. hint: 'git pull ...') before pushing again. hint: See the 'Note about fast-forwards' in 'git push --...
1class Solution2{3public:4int count(int n, int x)5 {6int cnt = 0, k;7for (int i = 1; k = n / i; i *= 10)8 {9int high = k / 10; 10if (x == 0) 11 { 12if (high) 13 { 14 high--; 15 } 16else17 { 18break; 19 } 20 } 21 cnt += high * i; 22int c...
/* 题意： 给出数组A，有以下几个操作： 1: AND(opn, L, R):把区间[L, R]中的元素A[i]改为A[i] & opn;;;;;; 2: OR(opn, L, R) :把区间[L, R]中的元素A[i]改为A[i] | opn;;;;;;; 3: XOR(opn, L, R):把区间[L, R]中的元素A[i]改为A[i] ^ opn;;;;;;; 4: SUM(L, R) :对区间[L, R]中的元素求和；；；； -------------------------------------------------------------------------------- 线段树区间修改的题目： ----------------...
Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.Example: Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99]) Credits:Special thanks to @memoryless for adding this problem and creating all test cases.1234567891011121314151617181920212223242526classSolution: defcountNumber...
Digital Counter time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Malek lives in an apartment block with 100 floors numbered from 0 to 99. The apartment has an elevator with a digital counter showing the floor that the elevator is currently on. The elevat...