题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1196 Lowest Bit Description Given an positive integer $A (1 \leq A \leq 100)$, output the lowest bit of $A.$ For example, given $A = 26$, we can write $A$ in binary form as $11010$, so the lowest bit of…

题目链接:http://code.hdu.edu.cn/game/entry/problem/show.php?chapterid=1&sectionid=2&problemid=22 题目意思:给出一个数,观察其二进制表示,从右往左看,记录遇到第一个出现1的位置pos,做2 ^ pos 的运算. 这几天杭电的告示:Exercise Is Closed Now!  再加上想用一些简单的题目来调剂一下,因此就做ACM Steps  吧. 用了递归的方法来做. #include <ios…

Lowest Common Multiple Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34980    Accepted Submission(s): 14272 Problem Description 求n个数的最小公倍数.   Input 输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数.  …

Problem Description Given an positive integer A (1 <= A <= 100), output the lowest bit of A. For example, given A = 26, we can write A in binary form as 11010, so the lowest bit of A is 10, so the output should be 2. Another example goes like this:…

Problem Description Given an positive integer A (1 <= A <= 100), output the lowest bit of A. For example, given A = 26, we can write A in binary form as 11010, so the lowest bit of A is 10, so the output should be 2. Another example goes like this:…

http://acm.hdu.edu.cn/showproblem.php?pid=2028 Problem Description 求n个数的最小公倍数.   Input 输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数.   Output 为每组测试数据输出它们的最小公倍数,每个测试实例的输出占一行.你可以假设最后的输出是一个32位的整数.   Sample Input 2 4 6 3 2 5 7   Sample Output 12 70   代码: #includ…

水题,原理是计算机组成原理中的负数的补码的求码.利用按位与可解. #include <iostream> using namespace std; int main() { int n; ) { cin >> n; ) break; cout <<(n&(-n))<<endl; } ; }…

第一次一次通过,逻辑太简单... #include<iostream> using namespace std; void main() { int n; while(cin>>n&&n!=0) { int temp=1; while(n%2==0) { temp*=2; n/=2; } cout<<temp<<endl; } }…

题目大意是给一个1-100的整数,要求首先转化成2进制,然后从最低位开始数起到不是0的位停止,输出这些位代表队额10进制数 #include <iostream> using namespace std; ]={,,,,,,}; int judge(int a)//判断输入数据的范围 { ) ; &&a>=) ; &&a>=) ; &&a>=) ; &&a>=) ; &&a>=) ;…

也称欧几里得算法 原理: gcd(a,b)=gcd(b,a mod b) 边界条件为 gcd(a,0)=a; 其中mod 为求余 故辗转相除法可简单的表示为: int gcd(int a, int b) { return b ==0? a:gcd( b, a% b); } 简洁而优雅. 例如:HDU 2028 Lowest Common Multiple Plus求n个数的最小公倍数. 最小公倍数=两数之积  /  最大公约数 这里防止中间过程溢出,先除以最大公约数,然后在求积. #includ…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1196 大水题   lowbit   的应用    (可以取出一个数二进制中的最后一个1.树状数组常用,Lowbit(x)=x&-x. ) 代码: #include <stdio.h> #include <string.h> #include <math.h> #include <algorithm> #include <iostream> #…

Problem Description 求n个数的最小公倍数. Input 输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数. Output 为每组测试数据输出它们的最小公倍数,每个测试实例的输出占一行.你可以假设最后的输出是一个32位的整数. Sample Input 2 4 6 3 2 5 7 Sample Output 12 70 #include <bits/stdc++.h> #define ll long long #define inf 100000000…

转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…

http://acm.hdu.edu.cn/showproblem.php?pid=2860 n个旅,k个兵,m条指令 AP 让战斗力为x的加入y旅 MG x旅y旅合并为x旅 GT 报告x旅的战斗力 几个需要注意的小问题是:已经被合并过的不能再参加合并这样就简化了并查集本身的复杂度只存在一层继承关系 一道简单的并查集题目,一开始不知道什么情况总是PE,格式上找不到错23333 后来参考了潘大神的代码,原来是最后回车的位置粗了偏擦23333 #include<stdio.h> #define i…

HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1908 Double Queue Description The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest, equipped with a modern computing environment provided by IBM Romania, and using mod…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4000 Recently, dobby is addicted in the Fruit Ninja. As you know, dobby is a free elf, so unlike other elves, he could do whatever he wants.But the hands of the elves are somehow strange, so when he cuts…

ZYB loves Score Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5268 Description One day,ZYB participated in the BestCoder Contest There are four problems. Their scores are 1000,1500,2000,2500 According to the r…

转载来自:http://www.cppblog.com/acronix/archive/2010/09/24/127536.aspx 分类一: 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029.1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093.1094.1095.1096.1097.1098.1106.1108.1157…

模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201 120…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4930 Fighting the Landlords Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 312    Accepted Submission(s): 100 Problem Description Fighting the…

题目链接: PKU:http://poj.org/problem?id=3481 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=1908 Description The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest, equipped with a modern computing environment provided by…

HDU ACM-Steps RECORD Chapter 1 Section 1 暖手题 1.1.1 A+B for Input-Output Practice (I) #include <stdio.h> int main() { int a,b; while(scanf("%d %d",&a,&b)==2) printf("%d\n",a+b); return 0; } 1.1.2 A+B for Input-Output Pract…

LCA(Lowest Common Ancesor) 1.基于二分搜索算法 预处理father[v][k]表示v的2的k次方层祖先,时间复杂度是O(nlogn),每次查询的时间复杂度是O(logn),预处理2k表的技巧在LCA之外也会用到.用链式前向星存图,相对vector邻接表要快. 一次dfs预处理出全部点的父亲结点,然后用2分思想,处理出每个点的2的k次方的父亲结点,对于LCA核心算法,首先把深度较深的移动到与另外一个水平,然后两个结点一起移动,但他们的父亲结点不同时,先上移动,最后返回当…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3045 It’s summer vocation now. After tedious milking, cows are tired and wish to take a holiday. So Farmer Carolina considers having a picnic beside the river. But there is a problem, not all the cows co…

题目链接: acm.hdu.edu.cn/showproblem.php?pid=1158 Employment Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6020    Accepted Submission(s): 2609 Problem Description A project manager wants…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1121 Complete the Sequence Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 451    Accepted Submission(s): 283 Problem Description You probably kno…

Employment Planning Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1158 Description A project manager wants to determine the number of the workers needed in every month. He does know the minima…

King's Pilots 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5644 Description The military parade will last for n days. There is a aerobatic flight shows everyday during the parade. On the ith day , Pi pilots are required. But the pilots are not wil…

King's Game 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5643 Description In order to remember history, King plans to play losephus problem in the parade gap.He calls n(1≤n≤5000) soldiers, counterclockwise in a circle, in label 1,2,3...n. The firs…