题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5494 题意:首先一个数t代表测试样例组数  然后两个数n,m 代表数列中有n个数  从中取m个数    给出两组数列  第一组数列代表Soda手中的数字,第二组数代表Beta手中的数,如果从beta手中的数中任意取m个数的和都小于从soda手中任意取出的m个数的和,输出yes否则输出no 题解:求soda手中最小的m个数的和  求beta手中最大m个数的和 #include<stdio.h> #i…

容斥原理+状压 Card Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1940    Accepted Submission(s): 907Special Judge Problem Description In your childhood, do you crazy for collecting the bea…

版权声明:来自: 码代码的猿猿的AC之路 http://blog.csdn.net/ck_boss https://blog.csdn.net/u012797220/article/details/35236457 简单二分图匹配.... Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1073   …

概率DP/数学期望/状压DP/容斥原理 kuangbin总结中的第14题 好神奇的做法……题解看kuangbin的代码好了…… //HDOJ 4336 #include<cstdio> #define rep(i,n) for(int i=0;i<n;++i) #define F(i,j,n) for(int i=j;i<=n;++i) #define D(i,j,n) for(int i=j;i>=n;--i) ; <<N]; int main(){ int n…

概率DP的题目,一直就不会做这类题目.dp[s]表示状态为s的时候再买多少张牌可以买全,表示的是一个期望值.dp[s] = 1 + P(empty) * dp[s] + P(had) * dp[s] + P(new) * dp[nst].从而可以解dp[s]. /* 4336 */ #include <iostream> #include <sstream> #include <string> #include <map> #include <queu…

水题,STL双端队列. /* 2319 */ #include <iostream> #include <cstdio> #include <cstring> #include <deque> #include <algorithm> using namespace std; int main() { int n, t; int i, j, k; deque<int> Q; #ifndef ONLINE_JUDGE freopen(&…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

Text Postcards always spoil my holidays. Last summer, I went to Italy. I visited museums and sat in public gardens. A friendly waiter taught me a few words of Italian. Then he lent me a book. I read a few lines, but I did not understand a word. Every…

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…

1.CardIO 识别 框架 GitHub 下载地址 配置 1.把框架整个拉进自己的工程,然后在 TARGETS => Build Phases => Link Binary With Libraries 里边分别加入下面这几个框架. Accelerate.framework MobileCoreServices.framework CoreMedia.framework AudioToolbox.framework AVFoundation.framework 2.在TARGETS =>…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…

Rectangles    HDOJ(2056) http://acm.hdu.edu.cn/showproblem.php?pid=2056 题目描述:给2条线段,分别构成2个矩形,求2个矩形相交面积. 算法:先用快速排斥判断2个矩形是否相交.若不相交,面积为0.若相交,将x坐标排序去中间2个值之差,y坐标也一样.最后将2个差相乘得到最后结果. 这题是我大一的时候做过的,当时一看觉得很水,写起来发现其实没我想的那么水.分了好几类情况没做出来.今天看了点关于判断线段相交的知识,想起了这题便拿来练…

前言 不说话,先猛戳 Ranklist 看我排名. 这是用 node 自动刷题大概半天的 "战绩",本文就来为大家简单讲解下如何用 node 做一个 "自动AC机". 过程 先来扯扯 oj(online judge).计算机学院的同学应该对 ACM 都不会陌生,ACM 竞赛是拼算法以及数据结构的比赛,而 oj 正是练习 ACM 的 "场地".国内比较有名的 oj 有 poj.zoj 以及 hdoj 等等,这里我选了 hdoj (完全是因为本地上…

题意:有n个点,问其中某一对点的距离最小是多少 分析:分治法解决问题:先按照x坐标排序,求解(left, mid)和(mid+1, right)范围的最小值,然后类似区间合并,分离mid左右的点也求最小值 POJ 3714 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> const int N = 1e5 + 5; const double INF = 1e…

题目传送门1 2 题意:从无序到有序移动的方案,即最后成1 2 3 4 5 6 7 8 0 分析:八数码经典问题.POJ是一次,HDOJ是多次.因为康托展开还不会,也写不了什么,HDOJ需要从最后的状态逆向搜索,这样才不会超时.判重康托展开,哈希也可. POJ //#include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<string> #include<stack…

Install application pavucontrol Run pavucontrol You will see the configuration about sound card and mic card. If can not solve it, Kmix shoud be used and combine it with pavucontrol…

补充新闻:程序员黑餐馆系统 给自己饭卡里充钱 ,技术是双刃剑,小心,小心! 前言 从M1卡的验证漏洞被发现到现今,破解设备层出不穷,所以快速傻瓜式一键破解不是本文的重点,年轻司机将从本文中获得如下技能. 如果你想简单快速的上手,你可以选择ACR122-like,Proxmark3等容易购买到的操作简单的设备,或者有个带有NFC功能并安装有安卓Mifare Classic Tool (MCT)软件的手机也是个不错的选择. 本文基于树莓派加RC522,PN532模块试验,如果你是刚入门的Geek爱好…

HDOJ 1728 http://acm.hdu.edu.cn/showproblem.php?pid=1728 BFS求最少转过的弯 #include <stdio.h> #include <string.h> #include <queue> using namespace std; ][]; ][]; //记录到当前点已经转过个弯,初始化为-1,检测该点是否已遍历 struct node { int x; int y; }now,temp; ,,,-}; ,,-,…

并查集   英文:Disjoint Set,即“不相交集合” 将编号分别为1…N的N个对象划分为不相交集合, 在每个集合中,选择其中某个元素代表所在集合. 常见两种操作: n       合并两个集合 n       查找某元素属于哪个集合 并查集实现的程序代码: int set[MAXN],rank[MAXN]; //set[i]=k表示i的父节点是k,rank[]存储树的深度. int FindSet(int x) { if(set[x]!=x) set[x]=FindSet(set[x])…

Fibonacci again and again Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3610    Accepted Submission(s): 1491 Problem Description 任何一个大学生对菲波那契数列(Fibonacci numbers)应该都不会陌生,它是这样定义的:F(1)=1;F(2)=2;…

Card Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 218    Accepted Submission(s): 86Special Judge Problem Description There are x cards on the desk, they are numbered from 1 to x. The score o…

[OpenJudge 3061]Flip The Card 试题描述 There are N× Ncards, which form an N× Nmatrix. The cards can be placed upwards or downwards. Now Acer is going to do some operations so that all the cards are placed upwards after the operations. In each operation,…

Problem Description Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want…

find your present (2) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 13802    Accepted Submission(s): 5194 Problem Description In the new year party, everybody will get a "special present"…

HDOJ 题目分类 //分类不是绝对的 //"*" 表示好题,需要多次回味 //"?"表示结论是正确的,但还停留在模块阶 段,需要理解,证明. //简单题看到就可以敲的 1000:    入门用: 1001:    用高斯求和公式要防溢出 1004:1012: 1013:    对9取余好了 1017:1021: 1027:    用STL中的next_permutation() 1029:1032:1037:1039:1040:1056:1064:1065: 10…

题目传送门 /* 求逆序数的四种方法 */ /* 1. O(n^2) 暴力+递推 法:如果求出第一种情况的逆序列,其他的可以通过递推来搞出来,一开始是t[1],t[2],t[3]....t[N] 它的逆序列个数是N个,如果把t[1]放到t[N]后面,逆序列个数会减少t[1]个,相应会增加N-(t[1]+1)个 */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std;…

题目传送门 题意:中文题面 分析:LCIS应用:设置b[]为a[]的反转,然后LCIS,若相等的是自己本身,则+1, 否则+2 代码: #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> using namespace std; const int MAXN = 2e2 + 10; const int INF = 0x3f3f3f3f; int a[MAX…

题目传送门 /* 题意:在一个矩阵里放炮台,满足行列最多只有一个炮台,除非有墙(X)相隔,问最多能放多少个炮台 搜索(DFS):数据小,4 * 4可以用DFS,从(0,0)开始出发,往(n-1,n-1)左下角走,x = cnt / n; y = cnt % n; 更新坐标, 直到所有点走完为止,因为从左边走到右边,只要判断当前点左上方是否满足条件就可以了 注意:当前点不能放炮台的情况也要考虑 g[x][y] == 'o'; 的错误半天才检查出来:) */ #include <cstdio> #…