Pie(浮点数二分)

Pie http://poj.org/problem?id=3122 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24544 Accepted: 7573 Special Judge Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of…

poj 1064 Cable master 判断一个解是否可行浮点数二分

poj 1064 Cable master 判断一个解是否可行浮点数二分题目链接: http://poj.org/problem?id=1064 思路: 二分答案,floor函数防止四舍五入代码: #include <iostream> #include <stdio.h> #include <math.h> #include <algorithm> using namespace std; const int maxn = 10005; const…

#AcWing系列课程Level-2笔记——4. 浮点数二分算法

浮点数二分算法编写浮点数二分,记住下面的思路,代码也就游刃有余了! 1.首先找到数组的中间值,mid=(left+right)>>1,区间[left, right]被划分成[left, mid]和[mid , right]. 2.然后通过check(mid)判断中间值是不是满足这个性质,保证落到区间里就对了,check是根据不同的题型编写的. 3.最后就能使用折半,缩小区间了,当认为区间已经很小的时候,比如<=10^-6,其实就找到了答案. 浮点数二分算法的核心浮点数二分的本质也是边…

Pie（二分）

ime Limit: 1000MS Memory Limit: 65536K Total Submissions: 8930 Accepted: 3235 Special Judge 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 variou…

POJ - 3122 Pie（二分）

http://poj.org/problem?id=3122 题意主人过生日,m个人来庆生,有n块派,m+1个人(还有主人自己)分,问每个人分到的最大体积的派是多大,PS每个人所分的派必须是在同一个派上切下来的. 分析二分答案,每次统计当前体积下能分配的人数. #include<iostream> #include<cmath> #include<cstring> #include<queue> #include<vector> #incl…

HUD 1969：Pie（二分）

Pie Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16056 Accepted Submission(s): 5654 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I…

HDU 1969 Pie【二分】

[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…

题解报告：hdu 1969 Pie（二分）

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…

POJ3122 Pie（二分）

题目链接:http://poj.org/problem?id=3122 题意:一堆人分蛋糕,每人蛋糕大小一样,求最大能分多少,蛋糕必须是整块整块的,不能两块拼一起.然后注意输入F个人最后要分F+1份. 思路:很简单很水,但是精度处理很恶心,wa了很多发,直接二分蛋糕的半径就行了 AC代码: #include<iostream> #include<vector> #include<cstdio> #include<algorithm> #include<…

hdoj 1969 Pie【二分】

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…

poj_3122:Pie（二分）

不算难的一道二分..各种玄学错误..eps小了T,大了WA..最后G++改成C++提交就AC了.. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<algorithm> using namespace std; ; //const double pi=acos(-1.0); const do…

POJ 3122 Pie【二分答案】

<题目链接> 题目大意: 将n个半径不一但是高度为1的蛋糕分给 F+1个人,每个人分得蛋糕的体积应当相同,并且需要注意的是,每个人分得的整块蛋糕都只能从一个蛋糕上切下来,而不是从几个蛋糕上东拼西凑而成.现在问每人分得蛋糕的体积是多少. 解题分析:就是普通的二分答案,但是要注意一下浮点型二分的结构,与整型二分略有不同. #include <cstdio> #include <cmath> #include <algorithm> using namespace…

POJ 3122 Pie（二分+贪心）

Pie Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22684 Accepted: 7121 Special Judge 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…

hdu 1969 Pie（二分查找）

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…

POJ 1064 Cable master 浮点数二分

http://poj.org/problem?id=1064 题目大意: 有N条绳子,他们的长度分别为Li,如果从它们中切割出k条长度相同的绳子的话,这K条绳子每条能有多长? 思路: 二分,设答案为mid=(L+R)/2, 如果以mid划分可以分割出不小于K条绳子,那么解>=mid, 否则解小于mid PS: 最后的输出坑死了.要保留两位整数且不进位....T T #include<cstdio> #include<cmath> const int MAXN= 10000+1…

UVA 12097 Pie（二分答案）

题目链接: 这道题虽然不是一道典型的二分答案题,但同样也可以用二分答案来做. 来二分面积为$area$的派,然后看看条件是否矛盾. 与其矛盾的便是$f+1$个人是否每个人都会有. 一个半径为$r$的派只能切出$floor(\pi r^2/x)$块. AC代码: #include<cstdio> #include<iostream> #include<cmath> #include<algorithm> using namespace std; const d…

HDU 2199 Can you solve this equation? 【浮点数二分求方程解】

Now,given the equation 8x^4 + 7x^3 + 2x^2 + 3x + 6 == Y,can you find its solution between 0 and 100; Now please try your lucky. Input The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines fol…

AcWing 开平方浮点数二分

#include<iostream> using namespace std; int main() { double x; cin>>x; ,r=x; ) //for(int i=0;i<100;i++) 可以不用精确位数,直接循环100次 { ; if(mid*mid>=x) r=mid; else l=mid; } printf("%lf\n",l); ; }…

HDU 1007 Quoit Design（二分+浮点数精度控制）

Quoit Design Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 47104 Accepted Submission(s): 12318 Problem Description Have you ever played quoit in a playground? Quoit is a game in which fla…

poj 1064 Cable master ，二分精度！！！

给出n根绳子,求把它们分割成K条等长的绳子的最大长度是多少? 二分用 for(int i=0; i<100; ++i) 取代 while(r-l>eps) 循环100次精度能达到1e-30,基本上能一般题目的精度要求. 而浮点数二分区间的话easy产生精度缺失导致死循环. #include<cstdio> double L[10000 + 10]; int n, k; int ok(double x) { int cnt = 0; for(int i=0; i<n;…

UVA1389 Hard Life[二分答案+最小割]

我真菜啊←地址求最大密度子图方案.密度=边数/点数假设E,V为最大密度子图的边数点数.则$\forall \rho$有$\rho \leqslant \frac{E}{V}$即$E- \rho V \geqslant 0$,也就是说密度最大时不等式去等号,密度要是小一些的话就应大于0,那可以二分密度寻找最大的密度,判断是否合法的方法:要看左边的等式有没有大于0,那可以把点看成权值是$-\rho$的点,边抽象成权值为$1$的点,且其向连接两端的点连有向边.然后做最大权闭合子图,保证边选上就得把…

poj 1064 高精度二分

Cable master Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 32191 Accepted: 6888 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to orga…