二分是一种很有效的减少时间开销的策略, 我觉得单列出二分专题有些不太合理, 二分应该作为一中优化方法来考虑 这几道题都是简单的使用了二分方法优化, 二分虽然看似很简单, 但一不注意就会犯错. 在写二分时, 会遇到很多选择题, 很多"分叉路口", 要根据实际情况选择合适的"路" HDU2199 Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/…

Pie Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10043    Accepted Submission(s): 3637 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I…

#include <iostream>#include <cmath>#include <iomanip>using namespace std; double pie[10005];int T, N, F;double PI = acos(-1.0); bool test(double x){ int cnt = 0; for(int i=1; i<=N; ++i) cnt += int(pie[i]/x); if(cnt >= F+1) return 1…

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…

Pie Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 67   Accepted Submission(s) : 34 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description My birthday is coming up and trad…

记得用PI=acos(-1)反三角函数求,用一次排序,然后二分和贪心 #include<iostream> #include<algorithm> #include<iomanip> #include<cmath> using namespace std; class Pie{ public: double r; double s; }; bool Check(Pie p[], int n, int f, double x); bool Cpm(Pie&am…

题目大意是要办生日Party,有n个馅饼,有f个朋友.接下来是n个馅饼的半径.然后是分馅饼了, 注意咯自己也要,大家都要一样大,形状没什么要求,但都要是一整块的那种,也就是说不能从两个饼中 各割一小块来凑一块,像面积为10的和6的两块饼(饼的厚度是1,所以面积和体积相等), 假设每人分到面积为5,则10分两块,6切成5.够分3个人,假设每人6.则仅仅能分两个了! 题目要求我们分到的饼尽可能的大! 仅仅要注意精度问题就能够了,一般WA 都是精度问题 运用2分搜索: 首先用总饼的体积除以总人数,得到…

Cable master Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2499    Accepted Submission(s): 936 Problem Description Inhabitants of the Wonderland have decided to hold a regional programming con…