Chinese people think of '8' as the lucky digit. Bob also likes digit '8'. Moreover, Bob has his own lucky number L. Now he wants to construct his luckiest number which is the minimum among all positive integers that are a multiple of L and consist of…

还是挺难的吧......勉强看懂调了半天 首先表达式可以写成 8(10^x -1)/9,题意为求一个最小的x使L | 8(10^x -1)/9 设d=gcd(L,8) L | 8(10^x -1)/9 <=>9L | 8(10^x -1) <=>9L/d | 10^x -1 (因为 9L/d 和 8/d 互质了 所以 9L/d 能整除(8/d)*(10^x-1)和 8/d 无关,所以可以去掉) <=>10^x 同余 1(mod 9L/d) 引理: 若a,n互质,则满足1…

The Luckiest Number 题目大意:给你一个int范围内的正整数n,求这样的最小的x,使得:连续的x个8可以被n整除. 注释:如果无解输出0.poj多组数据,第i组数据前面加上Case i: 即可. 想法:这题还是挺好的.我最开始的想法是一定有超级多的数输出0.然后...我就在哪里找啊找....其实这道题是一道比较好玩儿的数论题.我们思考:连续的8可用什么来表示出来?$\frac{(10^x-1)}{9}\cdot 8$.其实想到这一步这题就做完了.这题的精髓就在于告诉我们连续的连…

题意 Language:Default The Luckiest number Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7083 Accepted: 1886 Description Chinese people think of '8' as the lucky digit. Bob also likes digit '8'. Moreover, Bob has his own lucky number L. Now…

一.题目 Chinese people think of '8' as the lucky digit. Bob also likes digit '8'. Moreover, Bob has his own lucky number L. Now he wants to construct his luckiest number which is the minimum among all positive integers that are a multiple of L and consi…

Chinese people think of '8' as the lucky digit. Bob also likes digit '8'. Moreover, Bob has his own lucky number L. Now he wants to construct his luckiest number which is the minimum among all positive integers that are a multiple of L and consist of…

The Luckiest number Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 246264-bit integer IO format: %I64d      Java class name: Main Chinese people think of '8' as the lucky digit. Bob also likes digit '8'. Mor…

The Luckiest number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1163    Accepted Submission(s): 363 Problem Description Chinese people think of '8' as the lucky digit. Bob also likes digit '…

文章目录 题意 思路 特殊情况k=0 Source Code 1 Source Code 2 题意 给定一个数组和一个整数k,返回是否存在一个长度至少为2的连续子数组的和为k的倍数. 思路 和上一篇博客的思路基本一致. LeetCode subarray-sum-equals-k题解 所不同的是,子数组至少长度为2.因此需要一个缓冲区,延缓往Hash表中加数的操作. 另外,因为是和变成是k的倍数.利用同余的知识易得我们维护的前缀和是群ZkZ_kZk​中的和. ps.这里我犯了一个错误,没有考虑k…

该题没思路,参考了网上各种题解.... 注意到凡是那种11111..... 22222..... 33333.....之类的序列都可用这个式子来表示:k*(10^x-1)/9进而简化:8 * (10^x-1)/9=L * k (k是一个整数)8*(10^x-1)=9L*kd=gcd(9L,8)=gcd(8,L)8*(10^x-1)/d=9L/d*k令p=8/d q=9L/d p*(10^x-1)=q*k因为p,q互质,所以q|(10^x-1),即10^x-1=0(mod q),也就是10^x=1…