求出n的倍数m,要求m使用的不同数字最少,且最小. 一开始不知道怎么搜,因为不知道m由多少个不同的数字组成. 然后百度了一下,看到和数论有关. m可能使用的数字的个数可能为一个或者两个 a,aa,aaa....n+1个a, 将这些数%n,那么肯定有两个余数相等,抽屉原理.那么这两个数相减,得到的数肯定是n的倍数,且这两个数由a和0组成. 所以就知道怎么搜了,先搜m由一个数组成的情况,如果不存在,那么就搜两个数组成的情况,要注意全部搜完,因为题目要求m最小. #include <stdio.h>…

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 origina…

Given a non-negative integer num represented as a string, remove k digits from the number so that the new number is the smallest possible. Note: The length of num is less than 10002 and will be ≥ k. The given num does not contain any leading zero. Ex…

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…

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 an…

Problem: 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 w…

 FZU 2105  Digits Count Time Limit:10000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Practice Description Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are intege…

Revolving Digits Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25512    Accepted Submission(s): 5585 Problem Description One day Silence is interested in revolving the digits of a positive int…

Add Digits 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…

最近做的题记录下. 258. Add Digits 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. int addDigi…