HDU 2462 The Luckiest number】的更多相关文章

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 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…
HDU.1394 Minimum Inversion Number (线段树 单点更新 区间求和 逆序对) 题意分析 给出n个数的序列,a1,a2,a3--an,ai∈[0,n-1],求环序列中逆序对最少的个数. 前置技能 环序列 还 线段树的逆序对求法 逆序对:ai > aj 且 i < j ,换句话说数字大的反而排到前面(相对后面的小数字而言) 环序列:把第一个放到最后一个数后面,就是一次成环,一个含有n个元素序列有n个环序列. 线段树的逆序对求法:每个叶子节点保存的是当前值数字的个数.根…
hdu 6216 A Cubic number and A Cubic Number[数学] 题意:判断一个素数是否是两个立方数之差,就是验差分.. 题解:只有相邻两立方数之差才可能,,因为x^3-y^3=(x-y)(x^2+xy+y^2),看(x-y),就能很快想到不相邻的立方数之差是不可能是素数的:),,然后把y=x+1代入,得:p=3x^2+3x+1,所以可得判断条件为:①p-1必须能被3整除:②(p-1)/3必须能表示为两个相邻整数之积. #include<iostream> #inc…
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: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 980    Accepted Submission(s): 301 Problem Description Chinese people think of '8' as the lucky digit. Bob also likes digit '8…
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…
HDU 1394 Minimum Inversion Number(线段树求最小逆序数对) ACM 题目地址:HDU 1394 Minimum Inversion Number 题意:  给一个序列由[1,N]构成.能够通过旋转把第一个移动到最后一个.  问旋转后最小的逆序数对. 分析:  注意,序列是由[1,N]构成的,我们模拟下旋转,总的逆序数对会有规律的变化.  求出初始的逆序数对再循环一遍即可了. 至于求逆序数对,我曾经用归并排序解过这道题:点这里.  只是因为数据范围是5000.所以全…
JAVA+大数搞了一遍- - 不是很麻烦- - /* HDU 6093 - Rikka with Number [ 进制转换,康托展开,大数 ] | 2017 Multi-University Training Contest 5 题意: 求L,R之间的好数的个数,好数要求在某个d(>=2)进制下数位是0到d-1的 分析: d 进制下好数的个数为 d!-(d-1)! ,且满足 d^(d-1) <= K <= d^d 可知 若 N > d^d 则 1-N 包含前 d-1 个进制的所有…
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 '…
给定一个数,判断是否存在一个全由8组成的数为这个数的倍数 若存在则输出这个数的长度,否则输出0 /* 个人感觉很神的一道题目. 如果有解的话,会有一个p满足:(10^x-1)/9*8=L*p => 10^x-1=9*L*p/8 设m=9*L/gcd(L,8) 则存在p1使得 10^x-1=m*p1 => 10^x=1(mod m) 根据欧拉定理 10^φ(m)=1(mod m) 所以x一定是φ(m)的因数(这好像是某个定理). */ #include<iostream> #incl…
The kth great number Time Limit:1000MS     Memory Limit:65768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4006 Description Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a nu…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1394 Minimum Inversion Number                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)                                            Total Submission(s): 10…
Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17737    Accepted Submission(s): 10763 Problem Description The inversion number of a given number sequence a1, a2, ...,…
Balanced Number Problem Description A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some…
主题链接:http://acm.hdu.edu.cn/showproblem.php? pid=5062 Problem Description A positive integer x can represent as (a1a2-akak-a2a1)10 or (a1a2-ak−1akak−1-a2a1)10 of a 10-based notational system, we always call x is a Palindrome Number. If it satisfies 0<…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1562 Problem Description Happy new year to everybody!Now, I want you to guess a minimum number x betwwn 1000 and 9999 to let (1) x % a = 0;(2) (x+1) % b = 0;(3) (x+2) % c = 0;and a, b, c are integers be…
The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.   For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we w…
Minimum Inversion Number Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1394 Description The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai…
http://acm.hdu.edu.cn/showproblem.php?pid=1394  //hdu 题目   Problem Description The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj. For a given sequence of numbers a1, a2, ..…
Minimum Inversion Number Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1394 Appoint description:  System Crawler  (2015-03-30) Description The inversion number of a given number sequence a1, a…
Problem Description The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj. For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of…
K-th Nya Number Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ID: 3943 64-bit integer IO format: %I64d      Java class name: Main Arcueid likes nya number very much.A nya number is the number which has exactly…
Minimum Inversion Number [题目链接]Minimum Inversion Number [题目类型]最小逆序数 线段树 &题意: 求一个数列经过n次变换得到的数列其中的最小逆序数 &题解: 先说一下逆序数的概念: 在一个排列中,如果一对数的前后位置与大小顺序相反,即前面的数大于后面的数,那末它们就称为一个逆序. 一个排列中逆序的总数就称为这个排列的逆序数.逆序数为偶数的排列称为偶排列:逆序数为奇数的排列称为奇排列. 如2431中,21,43,41,31是逆序,逆序数…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1394 题意:给你一个0 — n-1的排列,对于这个排列你可以将第一个元素放到最后一个,问你可能得到的最多逆序对的个数 求出原始序列的逆序对的数目,然后进行n-1次将第一个元素放到最后一个的操作,每次操作后可以用O(1)复杂度求得新序列的逆序对数目 此题的关键点在于求出原始序列逆序对的数目,可以使用树状数组, 线段树, 归并等方法. 下面是树状数组的解法 #include <iostream> #i…
题目链接: 传送门 Minimum Inversion Number Time Limit: 1000MS     Memory Limit: 32768 K Description The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj. For a given sequence of numbe…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1394 解题报告:给出一个序列,求出这个序列的逆序数,然后依次将第一个数移动到最后一位,求在这个过程中,逆序数最小的序列的逆序数是多少? 这题有一个好处是输入的序列保证是0 到 n-1,所以不许要离散化,还有一个好处就是在计算在这个序列中比每个数大和小的数一共有多少个的时候可以在O(1)时间计算出来,一开始我没有意识到,还傻傻的用了两层for循环来每次都计算,当然这样果断TLE了.把一个数从第一个移…
http://acm.hdu.edu.cn/showproblem.php?pid=5898 题意:给出一个区间[l, r],问其中数位中连续的奇数长度为偶数并且连续的偶数长度为奇数的个数.(1<=L<=R<= 9*10^18) 思路:在比赛的时候只大概记得是怎么写的,但是就是不会写,虽然写过好几道可还是写不出来.回去后重新写又错了几次,主要前导零的情况没有考虑清楚.存的时候存长度和之前一位是偶数还是奇数和是否有前导零.然后根据条件判断才写了出来. #include <cstdio…