CS Course Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 593    Accepted Submission(s): 288 Problem Description Little A has come to college and majored in Computer and Science.Today he has lea…

A Math Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 817    Accepted Submission(s): 344 Problem Description You are given a positive integer n, please count how many positive integers…

CS Course Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Little A has come to college and majored in Computer and Science. Today he has learne…

CS Course Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Little A has come to college and majored in Computer and Science. Today he has learne…

Covering Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 187    Accepted Submission(s): 107 Problem Description Bob's school has a big playground, boys and girls always play games here after sch…

2017-08-31 16:19:30 writer:pprp 这道题快要卡死我了,队友已经告诉我思路了,但是做题速度很缓慢,很费力,想必是因为之前 的训练都是面向题解编程的缘故吧,以后不能这样了,另外这几天要学习一下动态规划, 先普及两个小知识点,这个点很简单很简单但是却卡了我很久很久, 现在作为模板记录下来: 1.将二进制数转化为十进制数 //转化为十进制 //test:ok ll twoTen(int a[]) { ll ans = ; ; i < N ; i++) { ans += (…

HDU 4847 Wow! Such Doge! pid=4847" style="">题目链接 题意:给定文本,求有几个doge,不区分大写和小写 思路:水题.直接一个个读字符每次推断就可以 代码: #include <stdio.h> #include <string.h> char c; char a[5]; int main() { a[5] = '\0'; int ans = 0; while ((c = getchar()) != E…

Problem Description Nike likes playing cards and makes a problem of it. Now give you n integers, ai(1≤i≤n) We define two identical numbers (eg: 2,2) a Duizi, and three consecutive positive integers (eg: 2,3,4) a Shunzi. Now you want to use these inte…

Problem Description Monkey A lives on a tree, he always plays on this tree. One day, monkey A learned about one of the bit-operations, xor. He was keen of this interesting operation and wanted to practise it at once. Monkey A gave a value to each nod…

Problem Description Bob's school has a big playground, boys and girls always play games here after school. To protect boys and girls from getting hurt when playing happily on the playground, rich boy Bob decided to cover the playground using his carp…

HDU 6188 Duizi and Shunzi 链接:http://acm.hdu.edu.cn/showproblem.php?pid=6188 思路: 签到题,以前写的. 实现代码: #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map&g…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6188 题意 有n个数字,每个数字小于等于n,两个相同的数字价值为1,三个连续的数字价值为1 .问这n个数字的最大价值是多少 思路 用map对这n个数字中每个数字出现的次数进行统计,用ans记录总价值.然后从数字1开始向后考虑,如果数字i出现的次数不小于2,那么ans加上数字i出现的次数除以二(两个相同的数字价值为1),然后让这个次数对2取模,接下来判断数字i还有没有(即:是否为1):如果为1,考虑数…

题意 给一颗\(n\)个节点的带点权的树,以\(1\)为根节点,\(q\)次询问,每次询问给出2个数\(u\),\(x\),求\(u\)的子树中的点上的值与\(x\)异或的值最大为多少 分析 先dfs一遍,得到dfs序,就可以将这个问题转化为求区间\([l,r]\)中的值与\(x\)异或值最大的经典问题, 就按dfs序建可持久化01字典树,查询的时候查区间\([in[u],out[u]]\)就行了,\(in[u]\)和\(out[u]\)存的分别是\(u\)的子树上的节点在dfs序上的起始位置和…

2017-08-31 16:48:00 writer:pprp 这个题比较容易,我用的是快速幂 写了一次就过了 题目如下: A Math Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description You are given a po…

Query on A Tree 时间限制: 8 Sec  内存限制: 512 MB提交: 15  解决: 3[提交][状态][讨论版] 题目描述 Monkey A lives on a tree. He always plays on this tree.One day, monkey A learned about one of the bit-operations, xor. He was keen of this interesting operation and wanted to pr…

B是一个看起来很KDT的题  但是因为KDT是n^1.5的所以t  而且因为KDT需要周期性的重建所以复杂度会更高 因为只有51种颜色 所以想当然的就去想了状态压缩 因为询问的区间范围 x一定是从1开始的 所以可以用cdq分治 用cdq分治去掉时间维度 按x排序 用线段树维护每个区间的状态(压缩颜色) 单点更新和区间查询 因为每次线段树做cdq的一部分之后 我们都要去除影响 但是不能rebuild 1 可以用一个last代表这个区间上一次使用线段树是什么时候 每次更新和查询的时候对他进行更新 2…

A.A Math Problem #include <bits/stdc++.h> using namespace std; typedef long long ll; inline ll read(){ ,f=;char ch=getchar(); ;ch=getchar();} +ch-';ch=getchar();} return x*f; } /***********************************************************/ ; ll a[max…

题意:就是一个集合分开,有两种区分 对子:两个相同数字,顺子:连续三个不同数字,问最多分多少个 解法:贪心,如果当前数字不构成顺子就取对子 /2,如果可以取顺子,那么先取顺子再取对子 #include <iostream> #include <stdio.h> #include <vector> #include <string.h> #include <map> using namespace std; struct Node{ int x,y…

题意:删除指定数字,求剩下数字的与或非值 解法:保存一下前缀和后缀 #include <iostream> #include <stdio.h> #include <vector> #include <string.h> #include <map> using namespace std; ],y1[],z1[]; ],y2[],z2[]; ]; map<int,int>Mp; int n,m; map<int,int>…

解法:发现..最多15个,那么..暴力一下啦 #include <iostream> #include <stdio.h> #include <vector> #include <string.h> #include <map> using namespace std; struct Node{ int x,y; }node[]; map<int,int>Mp; long long n; map<int,int>::ite…

上一场CF打到心态爆炸,这几天也没啥想干的 A Math Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description You are given a positive integer n, please count how many…

题目链接  Problems HDOJ上的题目顺序可能和现场比赛的题目顺序不一样, 我这里的是按照HDOJ的题目顺序来写的. Problem 1001 签到 #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) typedef long long LL; LL n, f[31]; int ans; int main(){ f[1] = 1LL; fo…

Query on A Tree Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others) Problem Description Monkey A lives on a tree, he always plays on this tree. One day, monkey A learned about one of the bit-operations, xor. He was…

Color it Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Do you like painting? Little D doesn't like painting, especially messy color paint…

题目分析:用一个字符串输入之后遍历每一位求和后,不断%10获取最后一位存储下来,逆序用对应的英文单词输出(注意输入为0的情况) #include<iostream> #include<string> using namespace std; ] = {"zero", "one", "two", "three", "four", "five", "six&…

Duizi and Shunzi Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Nike likes playing cards and makes a problem of it. Now give you n integers, a…

Covering Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Bob's school has a big playground, boys and girls always play games here after school.…

题目链接 题意 : 给你一棵树.树上的每个点都有点权.之后有若干次问询.每次问询给出一个节点编号以及一个整数 X .问你以给出节点为根的子树中哪个节点和 X 异或最大.输出这个值 分析 : 看到这种树上异或最值的问题 可以考虑使用 Trie 来解决 首先涉及到子树 我们可以利用 DFS 序来构造出每个根的子树 DFS 序有很好的性质.其子树的所有节点必定是序列中连续的一段 那么我们就可以对这个 DFS 序列建立可持久化 Trie 然后通过类似前缀和减法的方式得到问询节点子树表示的区间中 所有数组…

HDU - 4788 Hard Disk Drive Time Limit:1000MS   Memory Limit:32768KB   64bit IO Format:%I64d & %I64u [Submit]  [Go Back]  [id=47572" rel="nofollow">Status]   Description Yesterday your dear cousin Coach Pang gave you a new 100MB hard…

2017ACM/ICPC广西邀请赛 重现赛1005CS Course 题意 给一个数列a,每次询问去掉第p个的与和.或和,异或和. 题解 预处理前缀和.后缀和即可. 但是当时想都没想就写了个线段树.线段树就要注意不存在的区间,&操作返回1,其他返回0. 代码 const int N=201000; int n,p,a[N]; int add(int a,int b,int t){ if(t==0) return a&b; if(t==1) return a|b; return a^b; }…