FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 53352    Accepted Submission(s): 17788 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1009 FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 92493    Accepted Submission(s): 32082 Problem Description FatMouse prepared M…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1009 FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 93676    Accepted Submission(s): 32566 Problem Description FatMouse prepared M…

FatMouse' Trade Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description FatMouse prepared M pou…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 63198 Accepted Submission(s): 21342 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats guardin…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 41982    Accepted Submission(s): 13962 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pound…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1009 题目大意:肥鼠准备了 磅的猫粮,准备和看管仓库的猫交易,仓库里装有他最喜爱的食物 豆.仓库有 个房间.第 间房包含了 磅的 豆,需要 磅的猫粮.肥鼠不必为了房间中的所有 豆而交易,相反,他可以支付 % 磅的猫粮去交换得到 % 磅的 豆.这里, 表示一个实数.现在他将这项任务分配给了你:请告诉他,能够获得的 豆的最大值是多少. 题目要求:(输入)输入包含多组测试数据.对于每组测试数据,以包含了两…

FatMouse' Trade Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requ…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 36632    Accepted Submission(s): 12064 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats gu…

解题思路:一只老鼠共有m的猫粮,给出n个房间,每一间房间可以用f[i]的猫粮换取w[i]的豆,问老鼠最多能够获得豆的数量 sum 即每一间房间的豆的单价为v[i]=f[i]/w[i],要想买到最多的豆,一定是先买最便宜的,再买第二便宜的,再买第三便宜的 -----m的值为0的时候求得的sum即为最大值   所以先将v[i]从小到大排序.   FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/…

题意:就是老鼠要用猫粮换粮食,第i个房间一些东西,要用东西去换,可以不全换.问给定的猫粮最多能换多少粮食. 析:贪心算法.我们先算出来每个房间物品的平均价格是多少,肯定越低越好,并且如果能全换就全换,如果不能, 肯定是最后一次了,就把剩下全部换了,看看能换多少.求和. 代码如下: #include <iostream> #include <cstdio> #include <algorithm> #include <queue> #include <v…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 43381    Accepted Submission(s): 14499 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats gu…

版权声明:本文作者靖心,靖空间地址:http://blog.csdn.net/kenden23/.未经本作者同意不得转载. https://blog.csdn.net/kenden23/article/details/31418535 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favor…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 42953    Accepted Submission(s): 14336 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

题意: 一只老鼠用猫粮来换豆子,每个房间的兑换率不同,所以得尽量从兑换率高的房间先兑换.肥老鼠准备M磅猫粮去跟猫交易,让猫在warehouse中帮他指路,以找到好吃的.warehouse有N个房间,第i个房间包含J[i]磅豆子,且要求F[i]磅猫粮.肥老鼠不必交易房间里的所有豆子,相反,当他以F[i]*a% 磅猫粮交换,就可以拿到J[i]*a%磅豆子,这里a是一个实数.现在,他准备把作业分配给你:他能获得最大的豆子数是多少.输入:非负整数M和N,紧接着有N行,每行有2个正数,分别是J[i]和F[…

链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1109 FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. The warehouse has N rooms. The i-th room contains J…

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FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 44470    Accepted Submission(s): 14872 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1257 虽然分类是dp感觉还是贪心 比较水 #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; +; int d[maxn]; //d数组存储一套系统的目前的发射的最小的高度…

题意  输入n个老鼠的体重和速度   从里面找出最长的序列  是的重量递增时速度递减 简单的DP  令d[i]表示以第i个老鼠为所求序列最后一个时序列的长度  对与每一个老鼠i  遍历全部老鼠j  当(w[i] > w[j]) && (s[i] < s[j])时  有d[i]=max(d[i],d[j]+1)  输出路径记下最后一个递归即可了 #include<cstdio> #include<algorithm> using namespace std…

There are N schedules, the i-th schedule has start time s i  si and end time e i  ei (1 <= i <= N). There are some machines. Each two overlapping schedules cannot be performed in the same machine. For each machine the working time is defined as the…

题意:给你一系列个w,s.要你找到最长的n使得 W[m[1]] < W[m[2]] < ... < W[m[n]] and S[m[1]] > S[m[2]] > ... > S[m[n]] 即在这n个w,s中满足w[i]<w[j]&&s[i]>s[j],要求:体重严格递增,速度严格递减,原始顺序不定 首先将s从大到小排序,即顺数固定后转化为最长上升子序列问题. 案例: 6008 1300 6000 2100 500 2000 1000 40…

本题就先排序老鼠的重量,然后查找老鼠的速度的最长递增子序列,只是由于须要按原来的标号输出,故此须要使用struct把三个信息打包起来. 查找最长递增子序列使用动态规划法.主要的一维动态规划法了. 记录路径:仅仅须要记录后继标号,就能够逐个输出了. #include <stdio.h> #include <algorithm> using namespace std; const int MAX_N = 1005; struct MouseSpeed { int id, w, s;…

CF 628C 题目大意:给定一个长度为n(n < 10^5)的只含小写字母的字符串,以及一个数d,定义字符的dis--dis(ch1, ch2)为两个字符之差, 两个串的dis为各个位置上字符的dis之和,求和给定的字符串的dis为d的字符串,若含有多个则输出任意一个,不存在输出-1 解题思路:简单贪心,按顺序往后,对每一个字符,将其变为与它dis最大的字符(a或者z),d再减去相应的dis, 一直减到d为0,剩余的字母则不变直接输出.若一直到最后一位d仍然大于0,则说明不存在,输出-1. /…

Uva 11729  Commando War (简单贪心) There is a war and it doesn't look very promising for your country. Now it's time to act. You have a commando squad at your disposal and planning an ambush on an important enemy camp located nearby. You have N soldiers…

题目大意:原题链接 相邻两个字母如果不同,则可以结合为前一个字母,如ac可结合为a.现给定一个字符串,问结合后最短可以剩下多少个字符串 解体思路:简单贪心 一开始读题时,就联想到之前做过的一道题,从后往前贪心(关键),假设dp[i]表示从第i个字符开始到末尾结合后最短可以剩下的字符串数目. 然后拿笔在纸上画了画,发现果然是正确的.最后只要输出dp[0]即可. 好开心,从读题到AC总共不超过10分钟,傻逼了,刚开始误写为dp[sz-1]=0; #include<bits/stdc++.h> us…

HDU 1078 FatMouse and Cheese ( DP, DFS) 题目大意 给定一个 n * n 的矩阵, 矩阵的每个格子里都有一个值. 每次水平或垂直可以走 [1, k] 步, 从 (0, 0) 点开始, 下一步的值必须比现在的值大. 问所能得到的最大值. 解题思路 一般的题目只允许 向下 或者 向右 走, 而这个题允许走四个方向, 所以状态转移方程为 dp(x, y) = dp(nextX, nextY) + arr(x, y); dp 代表在 x, y 的最大值. 由于 下一…

发工资咯: Time Limit: 2000/1000ms (Java/Others) Problem Description: 作为广财大的老师,最盼望的日子就是每月的8号了,因为这一天是发工资的日子,养家糊口就靠它了,呵呵 但是对于学校财务处的工作人员来说,这一天则是很忙碌的一天,财务处的小胡老师最近就在考虑一个问题:如果每个老师的工资额都知道,最少需要准备多少张人民币,才能在给每位老师发工资的时候都不用老师找零呢? 这里假设老师的工资都是正整数,单位元,人民币一共有100元.50元.10元…

Ruin of Titanic Time Limit: 2000/1000ms (Java/Others) Problem Description: 看完Titanic后,小G做了一个梦.梦见当泰坦尼克号撞到冰山时,自己也在大船上.情况十分危急,不过这个时候船才刚进水,距离船身完全沉没还有一定时间(假如救生的船足够的话可以顺利逃生). 假设大船上共有n个人,每个人的重量为W1,W2,....,Wn;现在有若干小船,每一只船最大载重为100,且每一艘小船规定最多只能载2人.你能帮焦急的船长算出最少…