题目链接 : https://cn.vjudge.net/problem/Gym-101194D 题目大意 : 给你n个冰激凌球,让你用这些冰激凌球去垒冰激凌,要求是下面的这一个必须是他上面一个的两倍大,一个冰激凌需要m个冰激凌,问你最多能造几个冰激凌. 具体思路: 贪心 + 二分 ,首先对输入的n个冰激凌球进行排序,从小到大开始,然后对答案进行二分,从当前的开始,找第一个满足的,然后找完一个再继续找下一个,如果当前要检测的答案是k,那么就需要找k层满足情况的． AC代码: #include<b…

/** 题目:2016-2017 ACM-ICPC CHINA-Final Ice Cream Tower 链接:http://codeforces.com/gym/101194 题意:给n个木块,堆一个小塔要k个木块,满足相邻两个木块,上面的木块大小至少是下面的木块的两倍. 问最多可以堆出几个小塔. 思路:二分+贪心. 先二分最终可以堆出的小塔数x,然后确定了数量,就可以这样来贪心,把前x个木块作为所有小塔的顶部木块, 然后从剩下的木块中继续取前x小的,从小到大放在x个小塔的次顶部,然后继续.…

题目: Mr. Panda likes ice cream very much especially the ice cream tower. An ice cream tower consists of K ice cream balls stacking up as a tower. In order to make the tower stable, the lower ice cream ball should be at least twice as large as the ball…

2017-08-18 21:53:38 writer:pprp 题意如下: Problem D. Ice Cream Tower Input file: Standard Input Output file: Standard Ouptut Time limit: seconds Mr. Panda likes ice cream very much especially the ice cream tower. An ice cream tower consists of K ice crea…

题目链接: http://codeforces.com/gym/101194/attachments https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5922 题意: 给出 $N$ 个冰淇淋球,第 $i$ 个冰淇淋球大小为 $B_i$,现在已知冰淇淋球堆叠起来可组成一个冰淇淋. 对于上下相邻的两个冰淇淋球,只有上面的那个大小不…

被一道数位DP折磨得欲仙欲死之后,再做这道题真是如同吃了ice cream一样舒畅啊 #include<bits/stdc++.h> using namespace std; #define rep(i,a,b) for(int i=a;i<=b;++i) #define ms(arr,a) memset(arr,a,sizeof arr) #define debug(x) cout<<"< "#x" = "<<x&l…

题目大意: 对于给出的n个冰激凌球的大小,满足下面的球的大小是上一个的至少2倍,对于给出的k(由k的冰激凌球才能算作一个冰激凌塔),问n个冰激凌球可以最多堆出多少个高度为k的冰激凌塔 题目分析: 对于n个冰激凌球,显然我们得知可以堆出的高度为k的塔的数量在0~[n / k]之间,这里可以通过二分遍历每一种可能,初始时二分边界l==0,r==[n / k],每次取中间值mid=(l+r)/ 2,判断mid高度为k的塔能否堆出,如果可以则尝试mid为更大,否则则尝试mid为更小时,不断二分尝试mid…

题目链接: http://codeforces.com/gym/101194/attachments https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5923 题意: 现有 $N$ 支队伍参加比赛,只有一个队伍能获胜.给出每个队伍一个赔率 $A_i:B_i$,你往这个队投 $x$ 元,若该队获胜你可得到 $\frac{A_i+…

题解:二分加贪心,,,二分答案,然后进行判断,判断的时候,首先给每一组配一个最大的球,然后在向每一组里面填球,注意填球的时候要按组进行,每一组先填一个,然后更新每一组内的最小值,方便下一次寻找. #include<bits/stdc++.h> using namespace std; typedef long long ll; ; ll arr[N]; ll b[N]; bool mark[N]; int n,m; bool judge(int x){ ; ;i<=x;i++) b[i]…

题目:题目链接 思路:比赛中读错了题,题目要求选一个连续区间,却读成了随便选取几个柜台,英语要好好学啊,读懂题就很简单了,扫一遍就出结果了 AC代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <cstring> #include <vector> #incl…

题目: To encourage visitors active movement among the attractions, a circular path with ice cream stands was built in the park some time ago. A discount system common for all stands was also introduced. When a customer buys ice cream at some stand, he…

传送门 分析 这道题做了好长时间,题意就很难理解. 我们注意到这句话Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph 也就是说在最后的图中相同颜色的点构成一个连通图 也就是说如果1和2.3在不同的点共同出现,那么2和3一定不会在某个点共同出现.于是我们可以直接暴力dfs,在每个点对没有颜色的冰激凌贪心染最小的颜色. 需要注意有某种冰激凌没有出现的情况. trick 代码 #in…

http://codeforces.com/contest/805/problem/E [题意] 染色数是很好确定,最少染色数是max(si)(最小为1,即使所有的si都为0,这样是单节点树形成的森林需要1种颜色),关键是确定染色方案. 一开始没看出来树有什么用,但其实一句话很关键:Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph. 也就是说在原来的图中,含有相同的冰淇淋的点是…

传送门 Ice Cream Parlor Authored by dheeraj on Mar 21 2013 Problem Statement Sunny and Johnny together have M dollars they want to spend on ice cream. The parlor offers N flavors, and they want to choose two flavors so that they end up spending the whol…

ShareThis - By Vikas Verma Bluetooth low energy (BLE) is a feature of Bluetooth 4.0 wireless radio technology, aimed at new, principally low-power and low-latency, applications for wireless devices within a short range. As I discussed in my previous…

F - 贪心+ 二分 Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu   Description   A set of n<tex2html_verbatim_mark> 1-dimensional items have to be packed in identical bins. All bins have exactly the same length l<tex2html_ver…

Card Game Cheater Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1566    Accepted Submission(s): 822 Problem Description Adam and Eve play a card game using a regular deck of 52 cards. The rule…

A. Free Ice Cream 题目连接: http://www.codeforces.com/contest/686/problem/A Description After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer. At the start of the day they hav…

Heap Partition Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge A sequence S = {s1, s2, ..., sn} is called heapable if there exists a binary tree T with n nodes such that every node is labelled with exactly one element from the se…

E. Sonya and Ice Cream time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Sonya likes ice cream very much. She eats it even during programming competitions. That is why the girl decided that…

Sunny and Johnny together have M dollars which they intend to use at the ice cream parlour. Among N flavors available, they have to choose two distinct flavors whose cost equals M. Given a list of cost of N flavors, output the indices of two items wh…

题目链接: A. Free Ice Cream //#include <bits/stdc++.h> #include <vector> #include <iostream> #include <queue> #include <cmath> #include <map> #include <cstring> #include <algorithm> #include <cstdio> using…

题目传送门 /* 贪心/二分查找:首先对ai%=p,然后sort,这样的话就有序能使用二分查找.贪心的思想是每次找到一个aj使得和为p-1(如果有的话) 当然有可能两个数和超过p,那么an的值最优,每次还要和an比较 注意:不能选取两个相同的数 反思:比赛时想到了%p和sort,lower_bound,但是还是没有想到这个贪心方法保证得出最大值,还是题目做的少啊:( */ #include <cstdio> #include <algorithm> #include <cst…

Ice cream coloring 题解: 这个题目中最关键的一句话是, 把任意一种类型的冰激凌所在的所有节点拿下来之后,这些节点是一个连通图(树). 所以就不会存在多个set+起来之后是一个新的完全图. 所以只要直接去做就好了. 对于每个节点来说,染色. 代码: #include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freope…

The link to problem:Problem - D - Codeforces   D. Range and Partition  time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output Given an array a of n integers, find a range of values [x,y] (x≤y…

熊猫先生非常喜欢冰淇淋,尤其是冰淇淋塔.一个冰淇淋塔由K个冰淇淋球堆叠成一个塔.为了使塔稳定,下面的冰淇淋球至少要有它上面的两倍大.换句话说,如果冰淇淋球从上到下的尺寸是A0, A1, A2,···,AK 1,那么A0×2 ≤ A1, A1 × 2 ≤ A2,等等. 有一天,熊猫先生在街上走着,发现一家卖冰淇淋球的商店.冰淇淋球共有N个,大小分别为B0.B1.B2.··.BN−1.潘达先生想知道这些球最多能做出的冰淇淋塔数量. Input 输入的第一行给出了测试用例的数量,接下来是T , T组测…

对那个树进行dfs,在动态维护那个当前的冰激凌集合的时候,显然某种冰激凌仅会进出集合各一次(因为在树上形成连通块). 于是显然可以对当前的冰激凌集合贪心染色.暴力去维护即可.具体实现看代码.map不必要. #include<cstdio> #include<set> #include<vector> #include<map> #include<algorithm> using namespace std; map<int,bool>…

Description standard input/outputStatements Valera has only one electrical socket in his flat. He also has m devices which require electricity to work. He's got n plug multipliers to plug the devices, the i-th plug multiplier has ai sockets. A device…

题目链接 set维护最小值贪心, 刚开始用树的直径+单调队列没调出来... #include <iostream>#include <cstdio> #include <set> #define REP(i,a,n) for(int i=a;i<=n;++i) #define x first #define y second using namespace std; typedef pair<int,int> pii; , INF = 0x3f3f3f…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1383 题意:在平面图中,有一条河,用直线y=k表示,河上面(y>k)的都是敌方区域,y<k的都是水,现在有s个狙击手在水中不知道他们的位置:有n个敌军的士兵,已知他们的坐标: 狙击手有一个射击的范围 D,现在要满足所有敌方士兵都在狙击手的射击范围内,而且还要离河的距离M尽量的远,其中M是所有狙击手离河的距离最近的那个:如果不能满足输出“impossible” 距离最大化,我们可以二…