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…

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…

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 <…

http://codeforces.com/contest/754/problem/D 题意: 给定几组区间,找k组区间,使得它们的公共交集最大. 思路: 在k组区间中,它们的公共交集=k组区间中右端点最小值-k组区间中左端点最大值.如果我们要区间大,那我们应该尽量让左端点小,右端点大. 先对区间按照左端点排序,然后用优先队列处理. 将区间按照左端点从小到大的顺序一一进队列,只需要进右端点即可,如果此时队列内已有k个数,则队首就是这k组区间的最小右端点,而因为左端点是从小到大的顺序进队列的,所以…

http://codeforces.com/contest/754/problem/D 给出n条线段,选出k条,使得他们的公共部分长度最大. 公共部分的长度,可以二分出来,为val.那么怎么判断有k条线段有共同的这个长度,而且选他们出来呢? 可以把右端点减去val - 1,那么以后就只需要k条线段至少有一个交点就可以了. 那么怎么确定这个交点呢? 我的做法是直接离散,然后暴力找出覆盖次数>=k的那个点. 复杂度好像有点高, log2e10 * nlogn #include <cstdio>…

题意:题目简化了就是要给你n个区间,然后让你选出k个区间  使得这k个区间有公共交集:问这个公共交集最大能是多少,并且输出所选的k个区间.如果有多组答案,则输出任意一种.   这题是用优先队列来处理区间问题的,感觉挺典型的所以记录下来.   首先,要知道 选取的k个区间的最大交集=区间右端点中的最小值-区间左端点中的最大值.那么,要求得这这么k个区间是公共交集最大,就创建一个最小堆的优先队列(只存放区间的右端点):然后按左端点从小到大(先将区间按左端点排序)将区间放入优先队列中.每当优先队列的大…

还记得lyf说过k=2的方法,但是推广到k是其他的话有点麻烦.现在这里采取另外一种方法. 先将所有线段按照L进行排序,然后优先队列保存R的值,然后每次用最小的R值,和当前的L来维护答案即可.同时,如果Q的size()比k大,那么就弹出最小的R. 具体见代码: #include <stdio.h> #include <algorithm> #include <string.h> #include <set> #include <vector> #i…

2个多小时,弱智了..(连A都做不对,就不要做D了(迷)) #include<bits/stdc++.h> #define lowbit(x) x&(-x) #define LL long long #define N 100005 #define M 1000005 #define mod 2147483648LL #define inf 0x7ffffffff using namespace std; inline int ra() { ,f=; char ch=getchar()…

时隔一个月重返coding…… 期末复习了一个月也不亏 倒是都过了…… 就是计组61有点亏 复变68也太低了 其他都还好…… 假期做的第一场cf 三道题 还可以…… 最后room第三 standing383简直人生巅峰…… 看楼上楼下都是两道题的 如果A题不错那么多估计能进前300了吧…… 这场倒是把之前两场的分加回来了 开头不错 这个假期争取紫名~ A.Lesha and array splitting 把给定的数组分割成几个区间 要求各个区间和不能为0 一开始没注意到分割之后的区间重新合成之…

A. Lesha and array splitting 水题模拟.(0:10) 题意:给你一个n个元素的数组,求能否把这个数组分成若干连续小段,使得每段的和不为0.如有多种解输出任意一个. 思路:搞一个前缀和,如果这个数列的和不为0那么就直接输出一个区间左为1右为n.如果数列和为0那么找一个点前面的和不为0且后面的和也不为0,直接输出这两个区间. int a[N],sum[N]; int main() { int n; while(~scanf("%d",&n)) { mem…

  Codeforces Round #267 (Div. 2) CF#267D D - Fedor and Essay D. Fedor and Essay time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After you had helped Fedor to find friends in the «Call of S…

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…

B. Fedor and New Game time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game…

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…

任意门 为了在漫长得飞行旅途中娱乐,Alex和Fedor发明了如下的一个简单的双人游戏.游戏是: 首先, Alex画一个有权无向图.该图中可能有多重边(多重边的权值可能相同或者不同). 然后,Fedor选取该图的一个生成树.如果该树是这个图的最小生成树,则Fedor获胜.否则,Alex获胜. 如果两棵树分别包含的边标号形成的两个集合不相同,我们认为这两棵树不同.两个集合A和B,如果至少存在一个元素在A中但不在B中,或者反之,至少存在一个元素在B中但不在A中,则我们认为这两个集合不同. 我们还要指…

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…

D. Fedor and Essay time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the E…

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…

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…

After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game «Call of Soldiers 3». The game has (m + 1) players and n types of soldiers in total. Players «Call of Soldiers 3» are numbered for…

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…

D. Fedor and Essay time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the E…

After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game «Call of Soldiers 3». The game has (m + 1) players and n types of soldiers in total. Players «Call of Soldiers 3» are numbered for…

题目链接:http://codeforces.com/contest/467/problem/B 题目意思:有 m + 1 个 player 和 n 种类型的 soldiers.每个player被赋予一个数xi,然后将xi 看成二进制数,规定第 j 位 如果为1,表示这个 player 有j 这种类型的soldiers.Fedor 是 第 m + 1 个player,问他能跟前面 m 个players 成为 friends 的 人数.成为friends 的条件是被比较的两个人的不同soldier…