Pie Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12138    Accepted Submission(s): 4280 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1969 Pie Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4513    Accepted Submission(s): 1819 Problem Description My birthday is coming up and trad…

Pie Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16554    Accepted Submission(s): 5829   My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of…

Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This…

Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This…

[分析] “虽然不是求什么最大的最小值(或者反过来)什么的……但还是可以用二分的,因为之前就做过一道小数型二分题(下面等会讲) 考虑二分面积,下界L=0,上界R=∑ni=1nπ∗ri2.对于一个中值x和一张半径r的饼来说,这张饼能够分出的最多分数显然是:⌊π∗r2x⌋,所以我们只需要求出∑ni=1⌊π∗ri2x⌋,判断其与f+1的关系即可 特别要注意的一点:π的大小!,因为r≤104r≤104,所以r2≤108r2≤108,又因为题目要求精确到三位小数,所以ππ就要精确到10-11,即π=3.1…

题目大意:n块馅饼分给m+1个人,每个人的馅饼必须是整块的,不能拼接,求最大的. 解题思路: 1)用总饼的体积除以总人数,得到每个人最大可以得到的V.但是每个人手中不能有两片或多片拼成的一块饼. 代码如下: /* * 1969_2.cpp * * Created on: 2013年8月14日 * Author: Administrator */ #include <stdio.h> #include <math.h> #include <string.h> double…

1.题意:一项分圆饼的任务,一堆圆饼共有N个,半径不同,厚度一样,要分给F+1个人.要求每个人分的一样多,圆饼允许切但是不允许拼接,也就是每个人拿到的最多是一个完整饼,或者一个被切掉一部分的饼,要求你算出每人能分到的饼的体积最大值.输入数据依次给出,测试数据组数T,每组数据中,给出N,F,以及N个圆饼的半径.输出最大体积的数值,精确到小数点后四位. 2.分析:一看是这种输出就知道用二分写会很高效,这里对"能分出的最大体积值"进行二分.首先,这个值有界,最大值为总体积除以总人数的值,即Σ…

My birthday is coming up and traditionally I’m serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece…

二分答案+验证(这题精度卡的比较死) #include<stdio.h> #include<math.h> #define eps 1e-7 ; double a[ff]; double pi = acos(-1.0); int main() { int sb; scanf("%d", &sb); while (sb--) { int m, n; scanf("%d%d", &m, &n); int i, j; ; ;…

题目链接 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie.…

Pie Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3133    Accepted Submission(s): 1217 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I h…

<span style="color:#3333ff;">/* ----------------------------------------------------------------------------- author : Grant Yuan time : 2014.7.19 aldorithm: 01背包+卡精度 ------------------------------------------------------------------------…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6288 题意:给出a,b,k,n可满足(n^a)*(⌈log2n⌉)^b<=k ,求最大的n值三个正整数a,b,k(1≤a,b≤10,10^6≤k≤10^18) 题目思路:这类给数学式子求n的最大值,且数据量大且多的,考虑时间复杂度,我们采用二分法找出n 思路:很明显就是二分n,向上取整可以先预处理出2^62,然后直接循环找到b的底数j,处理n^a*j^b一开始我们用的是powl,但是被卡精度了,卡了…

HDOJ(HDU).2546 饭卡(DP 01背包) 题意分析 首先要对钱数小于5的时候特别处理,直接输出0.若钱数大于5,所有菜按价格排序,背包容量为钱数-5,对除去价格最贵的所有菜做01背包.因为这时候还剩下5块钱,直接买最贵的那个菜,就可以保证剩下来的钱数是最小的. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nma…

卡精度的任意模数fft模板题……这道题随便写个表就能看出规律来(或者说考虑一下实际意义),反正拿到这题之后,很快就会发现他是任意模数fft模板题.然后我就去网上抄了一下板子……我打的是最土的任意模数fft,就是fft7次的那种……(好像有很多方法的样子……)这种任意模数fft方法见http://blog.csdn.net/l_0_forever_lf/article/details/52886397这道题的具体做法见http://blog.csdn.net/qq_33229466/article…

描述 Given the coordinates of the vertices of a triangle,And a point. You just need to judge whether the point is in the Triangle. 输入 The input contains several test cases. For each test case, only line contains eight integer numbers , describing the c…

题意: 给定一个区间,a到b, n在区间内,有一个计算素数的公式,n*n+n+41,将n带进去可以得出一个数字.但是这个公式可能不准确,求出这个公式在这个区间内的准确率. 直接模拟就好了,不过要 注意 :直接对1到10000的数据打一个表,要不然会超时,在卡精度的时候加上一个1e-6或者1e-8,这东西有点说不清为什么,先记着,就像浮点型数据不能直接比较大小一样,比如: 定义一个浮点型的a和b,当判断a==b时,if(a-b<1e-6) #include<math.h> #include…

Pie Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12394    Accepted Submission(s): 4371 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I…

Pie Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7302    Accepted Submission(s): 2732 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I h…

题意:给了你n个蛋糕,然后分给m+1个人,问每个人所能得到的最大体积的蛋糕,每个人的蛋糕必须是属于同一块蛋糕的! 分析:浮点型二分,二分最后的结果即可,这里要注意圆周率的精度问题! #include<iostream> #include<stdio.h> #include<string.h> #include<math.h> using namespace std; #define pi acos(-1.0) #define pes 1e-8 ]; int…

题目链接:http://poj.org/problem?id=1064 题意:有n条绳子,长度为Li,现在从这n条绳子中切割出K条相等的绳子,问切割出来的这k条绳子的最大长度为多少: 二分判断即可: 但是本题的精度让人恶心: #include<iostream> #include<algorithm> #include<string.h> #include<stdio.h> #include<math.h> using namespace std…

Pie Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6320    Accepted Submission(s): 2383 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I h…

Cup Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8942    Accepted Submission(s): 2744 Problem Description The WHU ACM Team has a big cup, with which every member drinks water. Now, we know th…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=5652 Problem Description A long time ago there are no himalayas between India and China, the both cultures are frequently exchanged and are kept in sync at that time, but eventually himalayas rise…

http://acm.hdu.edu.cn/showproblem.php?pid=2546 呆呆. 饭卡 Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 12665    Accepted Submission(s): 4389 Problem Description 电子科大本部食堂的饭卡有一种很诡异的设计,即在购买之前判断余额.如…

http://acm.hdu.edu.cn/showproblem.php?pid=4750 题意: 定义f(u,v)为u到v每条路径上的最大边的最小值..现在有一些询问..问f(u,v)>=t的点对有所少对,注意(1,2)和(2,1)是不同的点对 分析: 原来最小生成树有一个很鬼畜的结论,那就是一个图的最小生成树中任意两个点的路径中的最大边一定最小.(妈蛋,完全不知道这个) 然后此题就变得很明朗了,用kruskal算法,加边的时候此边连接的两个集合的路径中的最大边就是这个边,存储下来,询问的时…

My birthday is coming up and traditionally I’m serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece…

题目大意:给你一个地图0代表可以通过1代表不可以通过.只要能从第一行走到最后一行,那么中国与印度是可以联通的.现在给你q个点,每年风沙会按顺序侵蚀这个点,使改点不可通过.问几年后中国与印度不连通.若一直联通输出-1. 题目思路:看数据这道题就是卡时间的,我们的基本思路是每当风沙侵蚀一个点,我们就进行一次广搜,看看图上下是否联通,我们应尽可能的去优化这个过程. 所以我们可以在遍历年的时候用二分查找: 若当年图可以上下联通,则继续向右查找 若当年图上下无法联通,则向左查找 剪枝: 为了省时间我们应该…

饭卡 Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 28562    Accepted Submission(s): 9876 Problem Description 电子科大本部食堂的饭卡有一种很诡异的设计,即在购买之前判断余额.如果购买一个商品之前,卡上的剩余金额大于或等于5元,就一定可以购买成功(即使购买后卡上余额为负),否则无…