(-￣▽￣)-* #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> using namespace std; int main() { int n; ]; while(~scanf("%d",&n)) { ;i<n;i++) { int m; scanf("%d",&m); w[i]=m*1.0; } s…

http://poj.org/problem?id=1862 题目大意: 有一种生物能两两合并,合并之前的重量分别为m1和m2,合并之后变为2*sqrt(m1*m2),现在给定n个这样的生物,求合并成一个的最小重量 思路: m1+m2 >=  2*sqrt(m1*m2) 所以每次取大的去合并,能变小. 直接优先队列就可以啦. #include<cstdio> #include<cmath> #include<queue> using namespace std;…

Stripies Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 18198   Accepted: 8175 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, bu…

原题链接:http://poj.org/problem?id=1862 简单题,贪心+优先队列主要练习一下stl大根堆 写了几种实现方式写成类的形式还是要慢一些... 手打的heap: 1: #include<cstdio> #include<cstdlib> #include<cmath> #include<iostream> class Solution{ public: ; int sz; double heap[Max_N]; inline void…

Stripies Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 10263   Accepted: 4971 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, bu…

最近学业上堕落成渣了.得开始好好学习了. 还有呀,相家了,好久没回去啦~ 还有和那谁谁谁... 嗯,不能发表悲观言论.说好的. 如果这么点坎坷都过不去的话,那么这情感也太脆弱. ----------------------------------------------------好好学习的分割线---------------------------------------------------- poj:http://poj.org/problem?id=3627 大意: 给一些东西的高度,…

Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, but the scientists had to invent an English name to apply for an international patent). The str…

每次合并最大的两个,优先级队列维护一下. 输出的时候%.3lf G++会WA,C++能AC,改成%.3f,都能AC. #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<algorithm> using namespace std; int n,s; priority_queue<double>Q; int main() { while…

题意:科学家发现一种奇怪的东西,他们有重量weight,如果他们碰在一起,总重变成2*sqrt(m1*m2).要求出最终的重量的最小值. 思路:每次选取质量m最大的两个stripy进行碰撞结合,能够得到最小的质量.所有只要维护一个优先队列就可以了 #include <iostream> #include <cstdio> #include <queue> #include <math.h> #include <cstring> #include…

POJ 1862 Stripies https://vjudge.net/problem/POJ-1862 题目:     Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, but the scientists had to invent an English n…

POJ 1488 题目大意:给定一篇文章,将它的左引号转成 ``(1的左边),右引号转成 ''(两个 ' ) 解题思路:水题,设置一个bool变量标记是左引号还是右引号即可 /* POJ 1488 Tex Quotes --- 水题 */ #include <cstdio> #include <cstring> int main() { #ifdef _LOCAL freopen("D:\\input.txt", "r", stdin); #…

题目链接: PKU:http://poj.org/problem?id=1862 ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=543 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian -…

Stripies 直接上中文了 Descriptions 我们的化学生物学家发明了一种新的叫stripies非常神奇的生命.该stripies是透明的无定形变形虫似的生物,生活在果冻状的营养培养基平板菌落.大部分的时间stripies在移动.当他们两个碰撞,会有新stripie生成,而旧的不见了.经过长期研究,他们发现新stripies的体重不等于消失的stripies的体重,而是:如果一个质量为m1和m2的stripies相撞,生成的stripies体重是2*sqrt(m1*m2) 现在,科学…

题目描述:http://poj.org/problem?id=1862 题目大意:你有n个数要合并,每两个数x,y合并后得到2*sqrt(x*y).求最后留下的一个数的最小值. 每合并一次,就会有数被开方,那么你越早合并的数被开放的次数越多,于是每次把最大的两个数合并即可.用到优先队列. 代码: #include<cstdio> #include<queue> #include<cmath> using namespace std; priority_queue<…

Stripies Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 14151   Accepted: 6628 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, bu…

题意: 有n个数,要把其中2个数进行2*根号(n1*n2)操作,求剩下最小的那个数是多少? 哭诉:看题目根本没看出来要让我做这个操作. 思路: 每次把最大的,次大的拿出来进行操作 用"优先队列"巧解,优先队列中剩下的那个就是题目要求求的答案. 解题代码: #include <iostream> #include <math.h> #include <algorithm> #include <queue> #include <cstd…

#include <string.h> #include <iostream> #include <queue> #include <stdio.h> using namespace std; struct product{ int deadline; int val; friend bool operator<(product n1,product n2) { return n1.val<n2.val; } }q; ]; int main()…

http://poj.org/problem?id=3614 有c头奶牛在沙滩上晒太阳,每头奶牛能忍受的阳光强度有一个最大值(max_spf) 和最小值(min_spf),奶牛有L种防晒霜,每种可以固定阳光强度在某一个值,每种的数量是cover[i] ,每头奶牛只能用一瓶防晒霜,问最多有多少头奶牛能在沙滩上晒太阳. 理解题意之后还是挺好做的. 首先确定的贪心策略是,在满足min_spf的条件下,尽量用spf小的用在max_spf大的奶牛身上,用一个最小堆维护max_spf的最小值即可. 先对奶牛…

题目链接: http://poj.org/problem?id=3614 Sunscreen Time Limit: 1000MSMemory Limit: 65536K 问题描述 to avoid unsightly burns while tanning, each of the c (1 ≤ c ≤ 2500) cows must cover her hide with sunscreen when they're at the beach. cow i has a minimum and…

题目链接:http://poj.org/problem?id=1456 Time Limit: 2000MS Memory Limit: 65536K Description A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx that is measured as an integral number of tim…

LINK 题意:给出1~n数字的排列,求变为递增有序的最小交换次数 思路:水题.数据给的很小怎么搞都可以.由于坐标和数字都是1~n,所以我使用置换群求循环节个数和长度的方法. /** @Date : 2017-07-20 14:45:30 * @FileName: LightOJ 1166 贪心 或 置换群 水题.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github…

Bridge over a rough river Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4143   Accepted: 1703 Description A group of N travelers (1 ≤ N ≤ 50) has approached an old and shabby bridge and wishes to cross the river as soon as possible. Ho…

Tian Ji -- The Horse Racing Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 12490   Accepted: 3858 Description Here is a famous story in Chinese history. That was about 2300 years ago. General Tian Ji was a high official in the country Q…

本文含有原创题,涉及版权利益问题,严禁转载,违者追究法律责任 本次是最后一篇免费的考试题解,以后的考试题目以及题解将会以付费的方式阅读,题目质量可以拿本次作为参考 本来半个月前就已经搞得差不多了,然后给一位神犇orz看(神犇orz都是很忙的!),就一直听取意见修修改改呀拖到了半个月之后,不过这也是为了能够做到完美吧 T1-apple-1s 第一题是一道贪心的水题 天数只有两天,不是今天吃就是明天吃,我们将 b[i]=min(x[i],y[i])定为基础开心值,也就是说不论哪天吃都至少可以得到这个…

http://poj.org/problem?id=3461 直接KMP就好.水题 #include<cstdio> #include<cstring> const int MAXN=10000+10; const int MAXM=1000000+10; char P[MAXN],T[MAXM]; int f[MAXN],n,m,ans; void getFail() { f[0]=f[1]=0; for(int i=1;i<n;i++){ int j = f[i]; wh…

地址 http://poj.org/problem?id=2431 题解 朴素想法就是dfs 经过该点的时候决定是否加油 中间加了一点剪枝 如果加油次数已经比已知最少的加油次数要大或者等于了 那么就剪枝 然而 还是TLE了 TLE代码 #include <iostream> #include <vector> #include <algorithm> #include <queue> using namespace std; vector<pair&l…

题目链接:http://poj.org/problem?id=2389 题目大意: 大数相乘. 解题思路: java BigInteger类解决 o.0 AC Code: import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); while (sc.h…

Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 110991   Accepted: 34541 Description Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical,…

487-3279 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 236746   Accepted: 41288 Description Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phras…

Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 99450   Accepted: 48213 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're as…