Hou Yi's secret Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1881    Accepted Submission(s): 450 Problem Description Long long ago, in the time of Chinese emperor Yao, ten suns rose into the…

题意: 给n个点,问最多有多少个相似三角形(三个角对应相等). 解法: O(n^3)枚举点,形成三角形,然后记录三个角,最后按三个角度依次排个序,算一下最多有多少个连续相等的三元组就可以了. 注意:在同一个坐标的两点只算一次,所以要判一下. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #includ…

直接6重循环就行了吧...判三角形相似直接从小到大枚举两向量夹角是否相等就行了.注意去重点跟三点共线就行了... #include<algorithm> #include<iostream> #include<cstring> #include<cstdlib> #include<fstream> #include<sstream> #include<bitset> #include<vector> #incl…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4082 题目大意: 给你n个点,问能最多构成多少个相似三角形. 用余弦定理,计算三个角度,然后暴力数有多少个,更新答案. 代码: #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #include <vector> #include <set>…

题意:射箭落在n个点,任取三点可构成一个三角形,问最大的相似三角形集(一组互相相似的三角形)的个数. 分析: 1.若n个点中有相同的点,要去重,题目中说射箭会形成洞,任选三个洞构成三角形,因此射在同一点只形成一个洞. 2.二进制枚举子集选出三个点,判断能否构成三角形. 3.因为边长可能为小数,因此用边长的平方判断相似. (1)将比较的两个三角形三边分别排序 (2)tmpx[0] / tmpy[0] = tmpx[1] / tmpy[1] = tmpx[2] / tmpy[2],即若满足tmpx[…

思路: 1.暴力枚举每种面值的张数,将可以花光的钱记录下来.每次判断n是否能够用光,能则输出0,不能则向更少金额寻找是否有能够花光的.时间复杂度O(n) 2.350 = 200 + 150,买350的道具可用一个150和200的代替,那么直接考虑200和150的道具即可.首先全部买150的道具,剩下的金额为x,可以看成t = x / 50张50的钱加上r = x % 50的钱.如果t <= n / 150,那么说明这些50的都可以和一个150的买200的道具,否则会剩下一些50的没法用.假设最后…

传送门 Fxx and string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 1007    Accepted Submission(s): 422 Description Problem DescriptionYoung theoretical computer scientist Fxx get a string which…

题目链接 题意:给平均成绩和科目数,求可能的最大学分和最小学分. 分析: 枚举一下,可以达到复杂度可以达到10^4,我下面的代码是10^5,可以把最后一个循环撤掉. 刚开始以为枚举档次的话是5^10,但是这个又不要求顺序,所以只是枚举个数就行了.. #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #includ…

jrMz and angles 题目链接: http://acm.hust.edu.cn/vjudge/contest/123316#problem/E Description jrMz has two types of angles, one type of angle is an interior angle of -sided regular polygon, and the other type of angle is an interior angle of -sided regula…

思路:直接暴力枚举区间[l,r]的整数值,然后max和min就可以了. AC代码: #pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<math.h> #include<algorithm> #inc…

Description Once upon a time Matt went to a small town. The town was so small and narrow that he can regard the town as a pivot. There were some skyscrapers in the town, each located at position x i with its height h i. All skyscrapers located in dif…

Calling Extraterrestrial Intelligence Again Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3993    Accepted Submission(s): 2097 Problem Description A message from humans to extraterrestrial in…

题目 这是一道可以暴力枚举的水题. //以下两个都可以ac,其实差不多一样,呵呵 //1: //4 wei shu #include<stdio.h> struct tt { ],b[],c[]; }e[]; int main() { ],mark[],yi,flag,a1,a2,a3,a4; while(scanf("%d",&n),n) { ;i<n;i++) { scanf("%s%s%s",e[i].a,e[i].b,e[i].c)…

Beautiful Now Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1876    Accepted Submission(s): 707 Problem Description Anton has a positive integer n, however, it quite looks like a mess, so he…

Safecracker Problem Description === Op tech briefing, 2002/11/02 06:42 CST ===  "The item is locked in a Klein safe behind a painting in the second-floor library. Klein safes are extremely rare; most of them, along with Klein and his factory, were de…

HDU 6638 - Snowy Smile 题意 给你\(n\)个点的坐标\((x,\ y)\)和对应的权值\(w\),让你找到一个矩形,使这个矩阵里面点的权值总和最大. 思路 先离散化纵坐标\(y\)的值 对\(n\)个点根据横坐标\(s\)进行排序 枚举横坐标,按顺序把点扔到线段树里,以离散化后\(y\)的\(id\)为下标\(pos\),存到线段树里 因为线段树可以在\(\log{n}\)的时间内插入数值,在\(O(1)\)的时间内查询当前区间最大子段和(线段树区间合并) \(node[…

题目:Click here 题意:给你n个点,有多少个正多边形(3,4,5,6). 分析:整点是不能构成正五边形和正三边形和正六边形的,所以只需暴力枚举四个点判断是否是正四边形即可. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define power(x) ((x)*(x)) using names…

Crazy Tank Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4562    Accepted Submission(s): 902 Problem Description Crazy Tank was a famous game about ten years ago. Every child liked it. Time f…

Description Math Olympiad is called “Aoshu” in China. Aoshu is very popular in elementary schools. Nowadays, Aoshu is getting more and more difficult. Here is a classic Aoshu problem: ABBDE __ ABCCC = BDBDEIn the equation above, a letter stands for a…

又一发吐血ac,,,再次明白了用函数(代码重用)和思路清晰的重要性. 11779687 2014-10-02 20:57:53 Accepted 4770 0MS 496K 2976 B G++ czy Lights Against Dudely Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1360    Accepted Subm…

abs Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given a number x, ask positive integer y≥2, that satisfy the following conditions:1. The absolute value of y - x is minimal2. To prime facto…

HDU 1557 权利指数 状态压缩 暴力 ACM 题目地址:HDU 1557 权利指数 题意:  中文题,不解释. 分析:  枚举全部集合,计算集合中的和,推断集合里面的团体是否为关键团队. 代码: /* * Author: illuz <iilluzen[at]gmail.com> * File: 1557.cpp * Create Date: 2014-06-28 14:47:58 * Descripton: brute force/ set */ #include <cstdio…

题目链接: hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5228 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=582&pid=1001 题解: 首先可以先考虑花色,把输入按照花色分组,对于每种花色同花顺的情况只有十种:(1,2,3,4,5)~(10,11,12,13,1),然后对每种花色的每种同花顺都对照一下,看看当前的牌有几张不在里面,这几张…

2017-08-25 20:08:54 writer:pprp 题目简述: • HDU 5616• n个砝码,可以放在天平左右两侧或不放• m次询问,每次询问是否可以测出给定重量• 1 ≤ n ≤ 20• 1 ≤ m ≤ 100 这道题采用枚举的思路的话实现起来还是有点困难的, 要实现的功能是对每个砝码进行处理,加到左边, 加到右边,或者是不加 看了大神的代码,感觉很巧妙, 设置了两个标记数组 vis1[2005], vis2[2005] 一个vis1用来记录当前已经可以实现的重量 另一个vis…

题意:求定 n 个数,求有多少对数满足,ai^bi = x. 析:暴力枚举就行,n的复杂度. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <c…

/* 将给定的一个字符串分解成ABABA 或者 ABABCAB的形式! 思路:暴力枚举A, B, C串! */ 1 #include<iostream> #include<cstring> #include<cstdio> #include<string> using namespace std; string str; ]; int main(){ int t; scanf("%d", &t); getchar(); while…

题目链接:http://acm.hnu.cn/online/?action=problem&type=show&id=12886&courseid=274 解题报告:输入4个数,要你判断用 + .- . * ./.四种运算能不能得到一个结果为24的式子,可以用括号. 解释一下测试的第四组样例:应该是6 / (1 - 3 / 4) 暴力枚举三种符号分别是什么,然后枚举这三种符号运算的顺序,然后枚举这四个数字的24种排列方式,时间是4^3 * 6 * 24 然后注意要用double型,…

题目链接 题意:中文题. 题解:暴力枚举. #include <iostream> #include <cstring> using namespace std; ; ; char num[MAXS]; int main(int argc, const char * argv[]) { while (cin >> num) { ; int len = (int)strlen(num); ; ; i < len; i++) { ') { > MINK) { M…

题目链接: http://www.codeforces.com/contest/666/problem/B 题意: 给你n个城市,m条单向边,求通过最短路径访问四个不同的点能获得的最大距离,答案输出一个满足条件的四个点. 题解: 首先预处理出任意两点的最短距离,用队列优化的spfa跑:O(n*n*logn) 现依次访问四个点:v1,v2,v3,v4 我们可以枚举v2,v3,然后求出v2的最远点v1,v3的最远点v4,为了保证这四个点的不同,直接用最远点会错,v1,v4相同时还要考虑次最远点来替换…

昨天梦到这道题了,所以一定要A掉(其实梦到了3道,有两道记不清了) 暴力枚举等的是哪张牌,将是哪张牌,然后贪心的判断就行了. 对于一个状态判断是否为胡牌,1-n扫一遍,然后对于每个牌,先mod 3, 如果还有剩余,就需要和i+1,i+2凑成顺子,减掉i+1,i+2的值就行了,然后 只要枚举到的i为负就是不可行,还有要枚举到n+2位,因为第n,n-1为可能 减了n+1,n+2但是没扫到. /******************************************************…