TJU Problem 2520 Quicksum】的更多相关文章

注意: for (int i = 1; i <= aaa.length(); i++) 其中是“ i <= ",注意等号. 原题: 2520.   Quicksum Time Limit: 0.5 Seconds   Memory Limit: 65536KTotal Runs: 2964   Accepted Runs: 1970 A checksum is an algorithm that scans a packet of data and returns a single…
注意代码中: result1 << " to " << result2 << ", PLAYER 1 WINS."<< endl; 和 result1 << " to " << result2 << ", PLAYER 1 WINS. "<< endl; 虽然只在WINS后只差一个空格,但会导致PE. 原题: 2101.   Bul…
下次不要被长题目吓到,其实不一定难. 先看输入输出,再揣测题意. 原文: 2548.   Celebrity jeopardy Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 1306   Accepted Runs: 898 It's hard to construct a problem that's so easy that everyone will get it, yet still difficult enough…
原题: 2857.   Digit Sorting Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 3234   Accepted Runs: 1704 Several players play a game. Each player chooses a certain number, writes it down (in decimal notation, without leading zeroes) and sorts t…
最重要的是找规律. 下面是引用 http://blog.sina.com.cn/s/blog_4dc813b20100snyv.html 的讲解: 做这题时,千万不要被那个图给吓着了,其实这题就是道简单的数学题. 首先看当m或n中有一个为2的情况,显然,只需要算周长就OK了.即(m+n-)*,考虑到至少其中一个为2,所以答案为2 *m或2*n,亦即m*n.注意这里保证了其中一个数位偶数. 当m,n≥3时,考虑至少其中一个为偶数的情况,显然,这种情况很简单,可以得出,结果为m*n,又可以和上面这种…
注意数据范围,十位数以上就可以考虑long long 了,断点调试也十分重要. 原题: 1065.   Factorial Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 6067   Accepted Runs: 2679 The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceiver…
注: 1. 对于double计算,一定要小心,必要时把与double计算相关的所有都变成double型. 2. for (int i = 0; i < N; i++)         //N 不可写为N - 1,否则当N为1时无法进行: 原题: 1100.   Pi Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 5683   Accepted Runs: 2317 Professor Robert A. J. Matthews…
注:对于每一横行的数据读取,一定小心不要用int型,而应该是char型或string型. 原题: 1090.   City hall Time Limit: 1.0 Seconds   Memory Limit: 65536KTotal Runs: 4874   Accepted Runs: 2395 Because of its age, the City Hall has suffered damage to one of its walls. A matrix with M rows an…
注意: int N; cin >> N; cin.ignore(); 同于 int N; scanf("%d\n",&N); 另:关于 cin 与 scanf: scanf是格式化输入,printf是格式化输出. cin是输入流,cout是输出流.效率稍低,但书写简便. 格式化输出效率比较高,但是写代码麻烦. 流输出操作效率稍低,但书写简便. cout之所以效率低,正如一楼所说,是先把要输出的东西存入缓冲区,再输出,导致效率降低. 缓冲区比较抽象,举个例子吧: 曾经…
题目链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2520 题意:有一个排列1~k,求第n个排列,其中n为 ,K(1≤K≤50000),S1, S2 ,…, Sk.(0≤Si≤K-i). 分析:这道题目乍看之下没有什么好的思路,k!太大了,但是仔细看一看就会发现n和康托展开式很类似 如果不知道康托展开的话请看:http://www.d…
https://www.luogu.org/problemnew/show/P4578 https://loj.ac/problem/2520 有点水的. 先转换成图论模型,即每个绿宝石,横坐标向纵坐标连边,权值为绿宝石要的数. 然后就变成了每个点,我按一下可以使得与它相连的边都+1/-1,问能否使图边权全部变成0. 其实你手玩一下的话你就会发现如果有解,你每次可以按这个点几下把一些边变成0,然后对于剩下的边我们当然可以用另一个端点把它变成零……持续下去你就会发现这么玩一定能使所有边都成0. 开…
Quicksum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16488   Accepted: 11453 Description A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will…
http://acm.fzu.edu.cn/problem.php?pid=2179 Problem 2179 chriswho Accept: 57    Submit: 136 Time Limit: 10000 mSec    Memory Limit : 327680 KB  Problem Description Chriswho很喜欢数字,特别喜欢一种数字,它能整除它的每一位数字(如果该位是0当做能整除),比如说126这个数字他就很喜欢,因为126%1=126%2=126%6=0.为…
Quicksum 时间限制(普通/Java):1000MS/3000MS          运行内存限制:65536KByte总提交:615            测试通过:256 描述 A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change…
Quicksum Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 102   Accepted Submission(s) : 33 Problem Description A checksum is an algorithm that scans a packet of data and returns a single number.…
我是菜鸟,我怕谁 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10445    Accepted Submission(s): 6041 Problem Description lin2144是一只小菜鸟,都是笨鸟先飞,lin2144想来个菜鸟先飞,他从0点出发一开始的飞行速度为1m/s,每过一个单位时间lin2144的飞行速度比上一…
POJ3094 Quicksum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18517   Accepted: 12712 Description A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the check…
2992.357000 1000 A+B Problem1214.840000 1002 487-32791070.603000 1004 Financial Management880.192000 1003 Hangover792.762000 1001 Exponentiation752.486000 1006 Biorhythms705.902000 1005 I Think I Need a Houseboat686.540000 1011 Sticks647.566000 1007…
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit cid=1006#status//H/0" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" style="font-family:Verdana,Arial,sans-serif; f…
最小树形图,測模版.... 2248.   Channel Design Time Limit: 1.0 Seconds   Memory Limit: 65536K Total Runs: 2199   Accepted Runs: 740 We need irrigate our farms, but there is only one source of water nearby. So we need build some water channels with minimum cost…
Quicksum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3516 Accepted Submission(s): 2579 Problem Description A checksum is an algorithm that scans a packet of data and returns a single number. T…
题目大意:http://acm.hdu.edu.cn/showproblem.php?pid=2520 我是菜鸟,我怕谁 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7698    Accepted Submission(s): 4505 Problem Description lin2144是一只小菜鸟,都是笨鸟先飞,lin2144…
Problem 5 # Problem_5.py """ 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? 什么是能够整…
求有限集传递闭包的 Floyd Warshall 算法(矩阵实现) 其实就三重循环.zzuoj 1199 题 链接 http://acm.zzu.edu.cn:8000/problem.php?id=1199 Problem B: 大小关系 Time Limit: 2 Sec  Memory Limit: 128 MBSubmit: 148  Solved: 31[Submit][Status][Web Board] Description 当我们知道一组大小关系之后,可判断所有关系是否都能成立…
Gson解析JSON字符串时出现了下面的错误: No-args constructor for class X does not exist. Register an InstanceCreator with Gson for this type to fix this problem. 解决的办法是把对应的Class改成静态类.…
C - NP-Hard Problem Crawling in process... Crawling failed Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 688C uDebug Description   Input   Output   Sample Input   Sample Output   Hint…
I joined the NodeJS online Course three weeks ago, but now I'm late about 2 weeks. I pay the codeschool yearly subscribed, but I have lost the track long time. I get more weight than expected. I like more and more my MacBook Pro I maybe go to the UST…
    Programming Contest Problem Types Hal Burch conducted an analysis over spring break of 1999 and made an amazing discovery: there are only 16 types of programming contest problems! Furthermore, the top several comprise almost 80% of the problems s…
题意: 给你一个数n,表示有n辆火车,编号从1到n,入站,问你有多少种出站的可能.    (题于文末) 知识点: ps:百度百科的卡特兰数讲的不错,注意看其参考的博客. 卡特兰数(Catalan):前几项为 : 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670- 令h(0)=1,h(1)=1,catalan数满足递推式:      h(n)= h(0…
2301: [HAOI2011]Problem b Time Limit: 50 Sec  Memory Limit: 256 MBSubmit: 4032  Solved: 1817[Submit][Status][Discuss] Description 对于给出的n个询问,每次求有多少个数对(x,y),满足a≤x≤b,c≤y≤d,且gcd(x,y) = k,gcd(x,y)函数为x和y的最大公约数. Input 第一行一个整数n,接下来n行每行五个整数,分别表示a.b.c.d.k Outp…