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…

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…

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…

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…

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 += (…

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…