[POJ2625][UVA10288]Coupons】的更多相关文章

Description Coupons in cereal boxes are numbered 1 to n, and a set of one of each is required for a prize (a cereal box, of course). With one coupon per box, how many boxes on average are required to make a complete set of n coupons? Input Input cons…
UVa10288 题目非常简单, 答案就是 n/n+n/(n-1)+...+n/1; 要求分数输出.套用分数模板.. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int maxn = 500; struct fraction { long long numerator; // 分子 long long…
题目: 题意: 一共n种不同的礼券,每次得到每种礼券的概率相同.求期望多少次可以得到所有n种礼券.结果以带分数形式输出.1<= n <=33. 思路: 假设当前已经得到k种,获得新的一种的概率是(n-k)/n,则对应期望是n/(n-k).求和得到步数期望是n/n+n/(n-1)+...+n/1=n*sum(1/i) (1<= i <= n).需要注意及时约分,用分数类模板. 程序: #include <cstdio> #include <cassert> #…
题目描述 “……在2002年6月之前购买的百事任何饮料的瓶盖上都会有一个百事球星的名字.只要凑齐所有百事球星的名字,就可参加百事世界杯之旅的抽奖活动,获得球星背包,随声听,更克赴日韩观看世界杯.还不赶快行动!” 你关上电视,心想:假设有n个不同的球星名字,每个名字出现的概率相同,平均需要买几瓶饮料才能凑齐所有的名字呢? 输入输出格式 输入格式: 整数n(2≤n≤33),表示不同球星名字的个数. 输出格式: 输出凑齐所有的名字平均需要买的饮料瓶数.如果是一个整数,则直接输出,否则应该直接按照分数格…
本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/转载请注明出处,侵权必究,保留最终解释权! Description All our characters have hobbies. The same is true for Fedor. He enjoys shopping in the neighboring supermarket. The…
D. Fedor and coupons time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output All our characters have hobbies. The same is true for Fedor. He enjoys shopping in the neighboring supermarket. The goo…
D. Fedor and coupons time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output All our characters have hobbies. The same is true for Fedor. He enjoys shopping in the neighboring supermarket. The goo…
Description Coupons in cereal boxes are numbered \(1\) to \(n\), and a set of one of each is required for a prize (a cereal box, of course). With one coupon per box, how many boxes on average are required to make a complete set of \(n\) coupons? Inpu…
uva10288:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1229 题意:有一种盒子,盒子一面会随机放一种卡片,现在要集齐n种卡片,问要买多少个盒子. 题解:假如说n==2,那么第一次买,肯定可以得到一种新的卡片,第二次买得到新的卡片的概率是1/2,也就是说你那买2包才能得到新的卡片,所以期望值是1+2…
Coupons and Discounts time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a…
B. Coupons and Discounts time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for…
最新亚马逊 Coupons 功能设置教程完整攻略! http://m.cifnews.com/app/postsinfo/18479 亚马逊总是有新的创意,新的功能.最近讨论很火的,就是这个 Coupons 的新功能,位于 Advertising 下面新增了 Coupons,如下图. 最新的查找亚马逊差评的方式,就看这篇! 亚马逊查找差评 ,最新再破解干货! 但是群里很多伙伴说自己的账号没看到 Coupons,小编都懂.因为小编的账号也没有(哭). 但是我们找到了新的路径,可以连接到 Coupo…
https://vjudge.net/problem/UVA-10288 大街上到处在卖彩票,一元钱一张.购买撕开它上面的锡箔,你会看到一个漂亮的图案. 图案有n种,如果你收集到所有n(n≤33)种彩票,就可以得大奖. 请问,在平均情况下,需要买多少张彩票才能得到大奖呢? 答案以带分数形式输出 例:当n=5时 思路简单,就是输出麻烦 #include<cstdio> #include<cstring> #include<cmath> #include<algori…
2590: [Usaco2012 Feb]Cow Coupons Time Limit: 10 Sec Memory Limit: 128 MB Submit: 349 Solved: 181 [Submit][Status][Discuss] Description Farmer John needs new cows! There are N cows for sale (1 <= N <= 50,000), and FJ has to spend no more than his bud…
2590: [Usaco2012 Feb]Cow Coupons Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 306  Solved: 154[Submit][Status][Discuss] Description Farmer John needs new cows! There are N cows for sale (1 <= N <= 50,000), and FJ has to spend no more than his bud…
P3045 [USACO12FEB]牛券Cow Coupons 71通过 248提交 题目提供者洛谷OnlineJudge 标签USACO2012云端 难度提高+/省选- 时空限制1s / 128MB 提交  讨论  题解 最新讨论更多讨论 86分求救 题目描述 Farmer John needs new cows! There are N cows for sale (1 <= N <= 50,000), and FJ has to spend no more than his budget…
题目链接 Coupons and Discounts 逐步贪心即可. 若当前位为奇数则当前位的下一位减一,否则不动. #include <bits/stdc++.h> using namespace std; #define rep(i,a,b) for(int i(a); i <= (b); ++i) ]; int n; int main(){ memset(a, , sizeof a); scanf("%d", &n); rep(i, , n) scanf…
UVA 10288 - Coupons option=com_onlinejudge&Itemid=8&page=show_problem&category=482&problem=1229&mosmsg=Submission+received+with+ID+13896541" target="_blank" style="">题目链接 题意:n个张票,每张票取到概率等价,问连续取一定次数后,拥有全部的票的期…
传送门1:http://www.usaco.org/index.php?page=viewproblem2&cpid=118 传送门2:http://www.lydsy.com/JudgeOnline/problem.php?id=2590 又挂了一道贪心,好烦啊. 这一题应该要想到“撤回”操作就好办了.假设现在已经没有优惠券了,那么如果你想以优惠价格买下一头牛,就要撤回以前的用优惠券买的牛,具体就是加回p[i] - c[i]就行了,可以理解为花这么多钱买一张优惠券.然后就是维护3个小根堆了.…
D. Fedor and coupons time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output All our characters have hobbies. The same is true for Fedor. He enjoys shopping in the neighboring supermarket. The goo…
time limit per test4 seconds memory limit per test256 megabytes inputstandard input outputstandard output All our characters have hobbies. The same is true for Fedor. He enjoys shopping in the neighboring supermarket. The goods in the supermarket hav…
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a number of year and he know…
Description Farmer  John  needs  new  cows! There  are  N  cows  for  sale (1 <= N <= 50,000), and  FJ  has  to  spend  no  more  than  his  budget  of  M  units  of  money (1 <= M <=$10^{14}$). Cow  i  costs $ P_i$  money (1 <=$ P_i$ <=…
[Usaco2012 Feb] Cow Coupons 一个比较正确的贪心写法(跑得贼慢...) 首先我们二分答案,设当前答案为mid 将序列按照用券之后能省掉的多少排序,那么我们对于两种情况 \(mid \leq k\) 全部取用券后的,取最小的\(mid\)个 排序后我们枚举分界点,对于右边\(p-c\)较大的,我们肯定考虑把k个券用给它们,就可以在左边取\(p\)最小的\(mid-k\)个,和右边的\(c\)最小的\(k\)个即可 这个我们用堆维护即可 #include<bits/stdc…
[USACO12FEB]牛券Cow Coupons(堆,贪心) 题目描述 Farmer John needs new cows! There are N cows for sale (1 <= N <= 50,000), and FJ has to spend no more than his budget of M units of money (1 <= M <= 10^14). Cow i costs P_i money (1 <= P_i <= 10^9), b…
LuoguP3045 [USACO12FEB]牛券Cow Coupons 果然我贪心能力还是太差了 ZR讲过的原题我回来对做法没有一丁点印象 有时候有这样一种题目 每个数有两种不同的价值 你可以选择价值低的,也可能花费一些神秘能力去获得价值高的 这时候我们直接贪心就可能会出现这种情况 当前最后解不是全局最优解 一般这种时候有两节决策, 要么DP 要么尝试进行可反悔的贪心 我们先按照所有牛的优惠后的价格排序,开一个小根堆 将前\(k\)个用优惠劵去买,很明显这可能是错误的 我们就将优惠券买的每一头…
P3045 [USACO12FEB]牛券Cow Coupons 贪心题.先选中 \(c_i\) 最小的 \(k\) 头牛,如果这样就超过 \(m\) ,直接退出,输出答案.否则考虑把后面的牛依次加入,替换前面用过券的牛.这里贪心得选择省钱最少的牛替换掉(这样影响最小,最有可能多买几头).加入牛的顺序按照 \(p\) 从小到大,因为如果换不掉就只能话 \(p\) 的钱去买 #include<bits/stdc++.h> using namespace std; #define int long…
题意: 每张彩票上印有一张图案,要集齐n个不同的图案才能获奖.输入n,求要获奖购买彩票张数的期望(假设获得每个图案的概率相同). 分析: 假设现在已经有k种图案,令s = k/n,得到一个新图案需要t次的概率为:st-1(1-s): 因此,得到一个新图案的期望为(1-s)(1 + 2s + 3s2 + 4s3 +...) 下面求上式中的级数: 令 则 所以得到一个新图案的期望为: 总的期望为: 这道题的输出很新颖,如果是分数的话,就要以分数形式输出,具体细节详见代码. #include <ios…
题意:给定n个优惠券,每张都有一定的优惠区间,然后要选k张,保证k张共同的优惠区间最大. 析:先把所有的优惠券按左端点排序,然后维护一个容量为k的优先队列,每次更新优先队列中的最小值,和当前的右端点, 之间的距离.优先队列只要存储右端点就好. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <…
好吧...想了半天想错了...虽然知道是贪心... 我们每次找没有被买的两种价格最小的牛,比较a = 当前差价最大的 + 当前优惠券价格最小的牛与b = 当前非优惠券价格最小的牛 所以...我们要 先维护两个小根堆,分别表示用优惠券买的牛的价格和不用优惠券买的牛的价格 还有个叫Recover的大根堆,表示当前几个用优惠券的那几头牛的差价(差价定义为非优惠价格与优惠价格的差值) a与b哪个小就买哪个... /*********************************************…