题目 1684: [Usaco2005 Oct]Close Encounter Time Limit: 5 Sec  Memory Limit: 64 MB Description Lacking even a fifth grade education, the cows are having trouble with a fraction problem from their textbook. Please help them. The problem is simple: Given a…

枚举分母,然后离他最近的分子只有两个,分别判断一下能不能用来更新答案即可 #include<iostream> #include<cstdio> #include<cmath> using namespace std; int a,b,aa,ab; double mx=10; void wk(int x,int y) { if(x*b==y*a) return; if(fabs((double)x/y-(double)a/b)<mx) { mx=fabs((dou…

1684: [Usaco2005 Oct]Close Encounter Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 387  Solved: 181[Submit][Status][Discuss] Description Lacking even a fifth grade education, the cows are having trouble with a fraction problem from their textbook. Pl…

http://www.lydsy.com/JudgeOnline/problem.php?id=1684 这货完全在考精度啊.. 比如奇葩 (llf)a/b*i (llf)(a/b*i)和(llf)(a/b)*i和(llf)(a/b)*(llf)i 这两货竟然不通????上边的能对,下边的就错了?? 噗. 全部都要..(llf)a/(llf)b*(llf)i..... 这样才不会错.. T_T 教训吸取了. #include <cstdio> #include <cstring>…

Description Lacking even a fifth grade education, the cows are having trouble with a fraction problem from their textbook. Please help them. The problem is simple: Given a properly reduced fraction (i.e., the greatest common divisor of the numerator…

Description As a reward for record milk production, Farmer John has decided to start paying Bessie the cow a small weekly allowance. FJ has a set of coins in N (1 <= N <= 20) different denominations, where each denomination of coin evenly divides th…

题目链接:http://begin.lydsy.com/JudgeOnline/problem.php?id=1333 题意: 有n种不同币值的硬币,并保证大币值一定是小币值的倍数. 每种硬币的币值为val,数量为cnt. 每个月你要给Bessie发金额为c的津贴(可以比c多,但不能少). 问你最多能发多少个月. 题解: 贪心. 贪心策略: (1)如果能恰好凑出c的钱,则应尽可能使用大币值的硬币. (2)如果不能恰好凑出,则应让花的冤枉钱尽可能少. 实现: 先按币值从大到小排序... (1)在保…

bzoj1745[Usaco2005 oct]Flying Right 飞行航班 题意: n个农场,有k群牛要从一个农场到另一个农场(每群由一只或几只奶牛组成)飞机白天从农场1到农场n,晚上从农场n到农场1,上面有c个座位,问最多可以满足多少只牛的要求.n≤10000,k≤50000,c≤100. 题解: 用类似贪心的方法做,现将每个农场出发的牛组织成链表.先求早上:当飞机到达每个农场时,先让到达的奶牛下机,接着如果飞机未满,则将其填满,之后枚举剩下的奶牛,如果它们的目的地比坐在飞机上面的奶牛目…

http://www.lydsy.com/JudgeOnline/problem.php?id=1685 由于每个小的都能整除大的,那么我们在取完大的以后(不超过c)后,再取一个最小的数来补充,可以证明这是最优的. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm>…

一开始直接 O( n² ) 暴力..结果就 A 了... USACO 数据是有多弱 = = 先sort , 然后自己再YY一下就能想出来...具体看code ----------------------------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #include<i…