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...
Different DigitsTime Limit: 10000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1138 Accepted Submission(s): 296Problem DescriptionGiven a positive integer n, your task is to find a positive integer m, which is a multiple of n, and that m contains the least number of different digits when represented in decimal. For example, number 1334 contains three d...
6> 【链接】5> 我是链接,点我呀:)6> 【题意】4> 在这里输入题意6> 【题解】4> //从大到小枚举第i(1..len1)位 //剩余的数字从小到大排序。 //看看组成的数字是不是小于等于b //如果是的话。 //说明第i位就是选这个数字了。 //接下来枚举下一位。6> 【代码】#include <bits/stdc++.h>#define ll long longusingnamespace std;constint N = 20;string s1,s2; int len1,len2; int bo[N],a[N],b[N]; vector<int> v,vans;bool get_ans(){...
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 = ...
Problem DescriptionGiven two positive integers n and k, you are asked to generate a new integer, say m, by changing some (maybe none) digits of n, such that the following properties holds:m contains no leading zeros and has the same length as n (We consider zero itself a one-digit integer without leading zeros.)m is divisible by kamong all numbers satisfying properties 1 and 2, m would be the on...
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
Given a non-empty string containing an out-of-order English representation of digits 0-9, output the digits in ascending order.Note:Input contains only lowercase English letters.Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as "abc" or "zerone" are not permitted.Input length is less than 50,000.Example 1:Input: "owoztneoer"Output: "01...
题意：给一个数字，每一次把它的最后一位拿到最前面，一直那样下去，分别求形成的数字小于，等于和大于原来数的个数。 SAM乱搞失败当然要先变SS了然后考虑每个后缀前长为n个字符，把它跟S比较就行了如果用后缀家族的话复杂度要加上log，本题会TLE吧求一个串S的每个后缀与另一个串T的最长公共前缀可以用扩展KMP！复杂度O(n+m)看课件吧 从1开始写真不容易以后再也不从1开始了，判断位置好麻烦好容易错next[i]=LCP(T[i,m],T)extend[i...
Given a non-empty string containing an out-of-order English representation of digits 0-9, output the digits in ascending order.Note:Input contains only lowercase English letters.Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as "abc" or "zerone" are not permitted.Input length is less than 50,000.Example 1:Input: "owoztneoer"Output: "01...
题目链接：D. Digits 思路：\(1e6 * 10\)的\(dp\)每次更新以k为结尾的最大值就好了（最开始的想法是正确的），有一点是因为是乘法，所以必须进行取log操作，让乘法变成加法，才不会爆。然后需要注意的是进行存放方案的操作有些许的麻烦。\(Code:\)#include<set> #include<iostream> #include<cstring> #include<cmath> #include<cstdio> #include<cstdlib> #include<map> #include<algorithm> #include<vector> #include<queue> #i...
题目Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.Example:Input: 38 Output: 2 Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.解法一思路不断地求新数字的 每个位 的和即可。代码class Solution {public int addDigits(int num) {int res = num;while(res / 10 != 0) {num = res;res = 0;while(num != 0) {res += ...
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...
descriptionGiven a positive integer N, return the number of positive integers less than or equal to N that have at least 1 repeated digit. ExampleInput: 1000 Output: 262 分析这道题目中 testcase 有错误，所以真的不知道哪些 ac 的答案是怎么做到的？ 犯错都是犯一样的错误～ count = 0 dup = [‘00‘, ‘11‘, ‘22‘, ‘33‘, ‘44‘, ‘55‘, ‘66‘, ‘77‘, ‘88‘, ‘99‘]for i in range(1, 1001):s = str(i)fo...
Given a non-negative integer N, find the largest number that is less than or equal to N with monotone increasing digits.(Recall that an integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= y.) Example 1:Input: N = 10 Output: 9 Example 2:Input: N = 1234 Output: 1234 Example 3:Input: N = 332 Output: 299 Note: N is an integer in the range [0, 10^9...
1. 问题描述Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.For example:Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.Follow up:Could you do it without any loop/recursion in O(1) runtime?Hint:A naive implementation of the above process is trivial. Could you come up with other methods?What are all...