传送门 zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 145    Accepted Submission(s): 49 Problem Description As one of the most powerful brushes in the world, zhx usually takes part…

Problem Description As one of the most powerful brushes in the world, zhx usually takes part in all kinds of contests. One day, zhx takes part in an contest. He found the contest very easy for him. There are n problems in the contest. He knows that h…

zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 455    Accepted Submission(s): 158 Problem Description As one of the most powerful brushes in the world, zhx usually takes part i…

zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 757    Accepted Submission(s): 282 Problem Description As one of the most powerful brushes in the world, zhx usually takes part in…

传送门 zhx's contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 575    Accepted Submission(s): 181 Problem Description As one of the most powerful brushes, zhx is required to give his juniors…

zhx's contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3835    Accepted Submission(s): 1255 Problem Description As one of the most powerful brushes, zhx is required to give his juniors n…

题目链接: hdu: http://acm.hdu.edu.cn/showproblem.php?pid=5187 bc(中文): http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=571&pid=1002 题解: 求(2^n-2)%p,题目看错,一天都没什么思路,冷静一下.. 代码: #include<iostream> #include<cstring> #include<cst…

作为史上最强的刷子之一,zhx的老师让他给学弟(mei)们出n道题.zhx认为第i道题的难度就是i.他想要让这些题目排列起来很漂亮. zhx认为一个漂亮的序列{ai}下列两个条件均需满足. 1:a1..ai是单调递减或者单调递增的. 2:ai..an是单调递减或者单调递增的. 他想你告诉他有多少种排列是漂亮的.因为答案很大,所以只需要输出答案模p之后的值. Input Multiply test cases(less than 10001000). Seek EOF as the end of…

题目分析如果n=1,答案是1,否则答案是2n−2. 证明:ai肯定是最小的或者最大的.考虑另外的数,如果它们的位置定了的话,那么整个序列是唯一的. 那么ai是最小或者最大分别有2n−1种情况,而整个序列单调增或者单调减的情况被算了2次,所以要减2. 要注意的一点是因为p>231,所以要用快速乘法.用法与快速幂相同.如果直接乘会超过long long范围,从而wa掉. 快速幂+快速乘法 #include<cstdio>#include<cstring>#include<c…

zhx's contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 448    Accepted Submission(s): 147 Problem Description As one of the most powerful brushes, zhx is required to give his juniors n p…

zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 324    Accepted Submission(s): 118 Problem Description As one of the most powerful brushes in the world, zhx usually takes part i…

In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3869    Accepted Submission(s): 1237 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project t…

HDOJ(HDU).3466 Dividing coins ( DP 01背包 无后效性的理解) 题意分析 要先排序,在做01背包,否则不满足无后效性,为什么呢? 等我理解了再补上. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 505 #define nn 505*100 using namespace std;…

HDOJ(HDU).2546 饭卡(DP 01背包) 题意分析 首先要对钱数小于5的时候特别处理,直接输出0.若钱数大于5,所有菜按价格排序,背包容量为钱数-5,对除去价格最贵的所有菜做01背包.因为这时候还剩下5块钱,直接买最贵的那个菜,就可以保证剩下来的钱数是最小的. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nma…

HDU 3732 Ahui Writes Word(多重背包) http://acm.hdu.edu.cn/showproblem.php? pid=3732 题意: 初始有N个物品, 每一个物品有cost[i]花费和val[i]价值, 你有m元钱, 如今问你最多能买多少总价值的物品? 当中N<=10W, m<=1W. 且cost[i]和val[i]都在[0,10]范围. 分析: 本题初看直接用01背包来做是直观的想法. 可是考虑到01背包的复杂度为O(N*m), 这么大的复杂度肯定不行. 然…

zhx's contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 851    Accepted Submission(s): 282 Problem Description As one of the most powerful brushes, zhx is required to give his juniors n p…

HDU 5234 题目大意:给定n,m,k,以及n*m(n行m列)个数,k为背包容量,从(1,1)开始只能往下走或往右走,求到达(m,n)时能获得的最大价值 解题思路:dp[i][j][k]表示在位置(i,j)有一个容量为k的背包所能获得的最大价值 决策:a[i][j]处的数是否选取 不选取: dp[i][j][k]= max(dp[i-1][j][k], dp[i][j-1][k]) 选取:首先要求k >=a[i][j],那么dp[i][j][k] = max(dp[i-1][j][k-w[i…

HDU 1284 钱币兑换问题(全然背包:入门题) http://acm.hdu.edu.cn/showproblem.php?pid=1284 题意: 在一个国家仅有1分,2分.3分硬币,将钱N (N<32768) 兑换成硬币有非常多种兑法. 请你编程序计算出共同拥有多少种兑法. 分析:基础的全然背包问题. 本题限制条件是: 金钱总数<=N. 本题目标条件是: 求构造方法数目. 令dp[i][j]==x 表示用前i种硬币构造j 美分共同拥有x种方法. 初始化:  dp为全0且dp[0][0]…

HDOJ(HDU).2602 Bone Collector (DP 01背包) 题意分析 01背包的裸题 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 1005 using namespace std; int v[nmax],w[nmax],dp[nmax]; int main() { //freopen("in…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2647 Problem Description Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewar…

HDU 1864 最大报销额 0-1背包 题意 现有一笔经费可以报销一定额度的发票.允许报销的发票类型包括买图书(A类).文具(B类).差旅(C类),要求每张发票的总额不得超过1000元,每张发票上,单项物品的价值不得超过600元.现请你编写程序,在给出的一堆发票中找出可以报销的.不超过给定额度的最大报销额. 这里要注意的是,只要总金额大于1000,或者每种类别的总金额大于600,或者这个发票中有其他类别物品,就不能报销 解题思路 这里就是简单的\(0-1\)背包了,主要的就是在于对于数据的预处…

先把收益为正数的处理掉:策略是挨个扫,扫n遍,碰到能买的就买,然后可以得到一个更新后的r 剩下的就看做是一个背包模型:物品(a,b)表示当背包体积>a时才能装下体积为b的该物品,问最多装几个 无序枚举所有物品显然错误,要选择先买哪个更优 对于两个物品(ai,bi),(aj,bj),有两种顺序 先买i,则至少需要max(ai,aj+bi)的体积 先买j,则至少需要max(aj,ai+bj)的体积 因为每买一件物品,剩余体积必定下降,为了达到某种状态,我们必须选择那种需要体积小的顺序 那么比较 aj…

有N道题.要求得到最少W分 给出N道题的:每道题用时T.分数V,应在且必须在L时刻提交才干得分 问得到W分所用的最少的时间 以L-T排序,然后做01背包就可以 #include "stdio.h" #include "algorithm" #include "string.h" using namespace std; struct Mark { int t,v,l,x; }mark[40]; int dp[300010]; bool cmp(M…

Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn't been so wealthy any more. The merchan…

题意:物品有三个属性,价格p,解锁钱数下线q(手中余额>=q才有机会购买该商品),价值v.钱数为m,问购买到物品价值和最大. 思路:首先是个01背包问题,但购买物品受限所以应先排序.考虑相邻两个物品1和2,假设先购买1再购买2,若…

Problem Description As one of the most powerful brushes, zhx is required to give his juniors n problems. zhx thinks the ith problem's difficulty is i. He wants to arrange these problems in a beautiful way. zhx defines a sequence {ai} beautiful if the…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5303 Problem Description There are n apple trees planted along a cyclic road, which is L metres long. Your storehouse is built at position 0 on that cyclic road. The ith tree is planted at position xi,…

The more, The Better Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6324    Accepted Submission(s): 3722 Problem Description ACboy很喜欢玩一种战略游戏,在一个地图上,有N座城堡,每座城堡都有一定的宝物,在每次游戏中ACboy允许攻克M个城堡并获得里面的宝物…

Magina Problem Description Magina, also known as Anti-Mage, is a very cool hero in DotA (Defense of the Ancient).If you have no idea of him, here is some brief introduction of his legendary antecedents:Twin sons to the great Prophet, Terrorblade and…

it's a  dp  difficult problem 试想如果我们遇见这样一道题,: 有n道题目,每道题有一个得分v和用时t: 我们要得够w分:用时最少  怎么做?? 这是一个裸奔的01背包 如果多一个规则, 每道题有一个开始时间,也就是开始做该题的时间不得少于一个时间st 现在怎么做? 我们应该把问题按照开始时间进行排序. 为什么? 用 st1 t1 与 st2 t2 来看 如果st1<st2 如果事情1先开始 那么所用时间ti1=st1+t1+(wait_time)t2 如果事情2先开…