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…

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…

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…

id=3122">[POJ 3122] Pie 分f个派给n+1(n个朋友和自己)个人 要求每一个人分相同面积 但不能分到超过一个派 即最多把一整个派给某个人 问能平均分的最大面积 二分平均面积 下界0 上界最大的一份派的面积 推断条件从大派開始分(保证尽量满足)假设能分出n+1份 这样的分法就合适 下界上移 最后输出下界就可以 注意二分推断上下界用esp 否则超时 从大到小分派是一种贪心策略 太小的派能够扔掉 但从小開始分有可能第一个派就分不出这么大 不排序也可遇到小派跳过 最后推断 代…

http://poj.org/problem?id=3122 题意:将n个圆柱体的不同口味的pie分给m个人,要求每个人分得的pie必须体积相同,且来自于一块pie(即:只分得一种口味的pie),求最多每个人可分得的体积. 思路:理解了题意就好做了,二分并注意精度. #include <stdio.h> #include <stdlib.h> #include <algorithm> #include <math.h> using namespace std…

Pie Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8594   Accepted: 3124   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 v…

Pie Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12985   Accepted: 4490   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…

Pie Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 109 Accepted Submission(s): 52   Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a…

二分查找是一个非常主要的算法,针对的是有序的数列,通过中间值的大小来推断接下来查找的是左半段还是右半段,直到中间值的大小等于要找到的数时或者中间值满足一定的条件就返回,所以当有些问题要求在一定范围内找到一个满足一些约束的值时就能够用二分查找,时间复杂度O(log n); 题目:http://acm.hit.edu.cn/hoj/problem/view?id=2651 由于题目有精度要求,对于浮点数小数点部分会有一定误差,所以能够选择将这些有小数部分的数值扩大e6倍,由于题目要求精确到e-3,之…

Pie Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 59   Accepted Submission(s) : 31 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description My birthday is coming up and trad…