[题目链接]:pid=5387">click here~~ [题目大意]给定一个时间点.求时针和分针夹角,时针和秒针夹角,分针和秒针夹角 模拟题,注意细节 代码: #include<bits/stdc++.h> using namespace std; inline int read(){ int c=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} whil…

题意: 给你一个格式为hh:mm:ss的时间,问:该时间时针与分针.时针与秒针.分针与秒针之间夹角的度数是多少. 若夹角度数不是整数,则输出最简分数形式A/B,即A与B互质. 解析: 先计算出总的秒数 S=hh∗3600+mm∗60+ss 由于秒钟每秒走1°, 所以当前时间,秒钟与12点的度数为 S%360 由于分针每秒走 0.1°, 既然已经计算出总秒数,那么当前时间,分针与12点的度数为 S/10%360 由于时针每秒走(1/120)°.那么当前时间.时针与12点的度数为 S/120%360…

题意:给一个时间,求三个时针之间的夹角,分数表示. 解法:算算算.统一了一下分母. 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h> #include<math.h> #include<limits.h> #include<time.h> #include<std…

题目传送门 /* 模拟:这题没啥好说的,把指针转成角度处理就行了,有两个注意点:结果化简且在0~180内:小时13点以后和1以后是一样的(24小时) 模拟题伤不起!计算公式在代码内(格式:hh/120, mm/120, ss/120) */ /************************************************ * Author :Running_Time * Created Time :2015-8-13 13:04:31 * File Name :H.cpp **…

HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011 亚洲北京赛区网络赛题目) Eliminate Witches! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 863    Accepted Submission(s): 342 Problem Description Kaname Mado…

算是最好写的一道题了吧,最近模拟没手感,一次过也是很鸡冻o(*￣▽￣*)o 注意事项都在代码里,没有跳坑也不清楚坑点在哪~ #include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<cmath> #include<sstream> #include<iostream> using namespace std; ; int h…

Clock Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 500    Accepted Submission(s): 176 Problem Description Given a time HH:MM:SS and one parameter , you need to calculate next time satisfying…

Clock Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 316    Accepted Submission(s): 215 Problem Description Give a time.(hh:mm:ss),you should answer the angle between any two of the minute.hou…

题 Description Give a time.(hh:mm:ss),you should answer the angle between any two of the minute.hour.second hand Notice that the answer must be not more 180 and not less than 0   Input There are $T$$(1\leq T \leq 10^4)$ test cases for each case,one li…

Thickest Burger Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 63    Accepted Submission(s): 60 Problem Description ACM ICPC is launching a thick burger. The thickness (or the height) of a piec…

Ugly Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0Special Judge Problem Description Everyone hates ugly problems.You are given a positive integer. You mu…

Football Games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 802    Accepted Submission(s): 309 Problem Description A mysterious country will hold a football world championships---Abnormal Cup…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4814 题目大意: 把一个正整数表示为φ进制, φ = (1+√5)/2 . 且已知: 1. φ + 1 = φ 2 ,所以有11(φ) = 100(φ),且要求11要转变为100 2. 2 * φ 2  = φ3 + 1 . 解题思路: 观察发现,2 = 10.01.所以对于一个数N,能够从N/2 推导得到,是一个模拟的过程.我比赛的时候居然用了高速幂... 代码: #include<cstdio>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5504 思路:模拟 代码: #include<stdio.h>//------杭电5504 #include<algorithm> #include<math.h> #include<iostream> using namespace std; int comp(const void*a, const void*b) { return *(int*)a - *(i…

http://acm.hdu.edu.cn/showproblem.php?pid=5071 对于每一个窗口,有两个属性:优先级+说过的单词数,支持8个操作:新建窗口,关闭窗口并输出信息,聊天(置顶窗口加单词),把优先级最高的窗口移到最前,把当前第i个窗口移到最前,把选择的窗口移到最前,把某个窗口置顶,把置顶窗口取消置顶...最后按优先级输出每个窗口的优先级以及单词数... 数据范围n<=5000-,恶心模拟 主要就是vector的操作,开始的时候没有置顶窗口记得把now置为0,注意每个参数范围…

http://acm.hdu.edu.cn/showproblem.php?pid=4902 出n个数,然后对这n个数进行两种操作: 如果是 1 l r x,则把 [l, r] 区间里面的每一个数都变为x: 如果是 2 l r x,则 比较 [l, r]区间里的数a_i和x的大小,如果a_i > x,把a_i变为a_i和x的最大公约数. 最后输出这n个数最终的值. 线段树可搞...但是没必要!... 线段树,每个结点多一个lazy表示该位置以下区间是否数字全相同,然后每次延迟操作,最后输出的时候…

http://acm.hdu.edu.cn/showproblem.php?pid=5186 题意是分别对每一位做b进制加法,但是不要进位 模拟,注意:1 去掉前置0 2 当结果为0时输出0,而不是全部去掉 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn=101; const int maxm=201; int n,b; char…

King's Phone 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5641 Description In a military parade, the King sees lots of new things, including an Andriod Phone. He becomes interested in the pattern lock screen. The pattern interface is a 3×3 square…

Isabella's Message Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4119 Description Isabella and Steve are very good friends, and they often write letters to each other. They exchange funny experiences, talk ab…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1014 Uniform Generator Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33120    Accepted Submission(s): 13137 Problem Description Computer simulati…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=2701 题意 有一个萤火虫会闪现 一个人 也会闪现 给出 这个人的起始位置 和他能够闪现的距离 然后依次给出萤火虫的坐标 这个人 每次都往萤火虫的坐标闪现 如果 这个人能够到达的距离方圆1个单位长度以内 萤火虫在里面 那么这个人就能够抓住萤火虫 就输出坐标 如果最后都没有抓住 那么 就是抓不住了 思路 只要模拟一下就好了 求这个人每次移动的坐标 学长点了我一下,,竟然是用相似三角形..初中的东西啊..…

题目链接: PKU:http://poj.org/problem? id=3344 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2414 Description Another boring Friday afternoon, Betty the Beetle thinks how to amuse herself. She goes out of her hiding place to take a walk around the living…

HDU 4884 TIANKENG's rice shop 题目链接 题意:模拟题.转一篇题意 思路:就模拟就可以.注意每次炒完之后就能够接单 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 1005; int T, n, t, k, m; struct Person { int t, p, num, ans; } p[N];…

HDU 4927 −ai,直到序列长度为1.输出最后的数. 思路:这题实在是太晕了,比赛的时候搞了四个小时,从T到WA,唉--对算组合还是不太了解啊.如今对组合算比較什么了-- import java.io.*; import java.math.*; import java.util.*; public class Main { public static void main(String[] args) { Scanner cin=new Scanner (new BufferedInput…

Time Zone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4337    Accepted Submission(s): 1278 Problem Description Chiaki often participates in international competitive programming contests. Th…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4925 解题报告:给你n*m的土地,现在对每一块土地有两种操作,最多只能在每块土地上进行两种操作,第一种是种苹果树操作,第二种是施肥操作,种苹果树操作可以使得该块地 长出一个苹果,施肥操作可以使得与这块土地相邻的土地的苹果产量变为原来的两倍,问可以得到的最多的苹果数量是多少? 例如一个4*4的土地,用1表示在该土地上做第一种操作,0表示在该土地上做第二种操作,可以得到最多苹果的操作如下: 0 1 0…

Problem Description Princess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of some value v, and all of them will come at a different time. Because the lobby is not large enough, Alisha can only let a…

Water problem 题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5867 Description If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3+3+5+4+4=19 letters used in total.If all the numbers from 1 to n (up to…

题意: 有两只兔子,一只在左上角,一只在右上角,两只兔子有自己的移动速度(每小时),和初始移动方向. 现在有3种可能让他们转向:撞墙:移动过程中撞墙,掉头走未完成的路. 相碰: 两只兔子在K点整(即处理完一小时走的路后),在同一点,兔子A和兔子B的方向互相交换一下. 向左转 : 两只兔子有自己的转向时间T,每隔T小时,它就会向左转, 但是相碰的优先级高于它,相碰之后就不处理左转问题了. 写模拟题还是很有趣的......练练手 #include <iostream> #include <a…

Problem Description As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes…