题目传送门 /* 题意:给n, k,然后允许给某一个数加上k的正整数倍,当然可以不加, 问你是否可以把这n个数变成1,2,3,...,n, 可以就输出Jerry, 否则输出Tom. 贪心:保存可能变成的值的方案数,当一个符合,其他所有可能方案减1 最大匹配 详细解释:http://blog.csdn.net/u012596172/article/details/40784773?utm_source=tuicool */ #include <cstdio> #include <algor…

Pearls Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other) Total Submission(s) : 22   Accepted Submission(s) : 12 Problem Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces…

开启DP之路 题目:http://acm.hdu.edu.cn/showproblem.php?pid=1069 描述一下: 就是给定N(N<=20)个方体,让你放置,求放置的最高高度,限制条件是:上面的物体的长度和宽度都小于下面物体. 觉得这道题目很经典: 于是贴下 .我们可以把一个方体看成六个,想到这里,然后求最高,方程很简单了 ,  if (a[j].x>a[i].x&&a[j].y>a[i].y)             ans=max(ans,a[j].h+a[…

最少拦截系统 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 38985    Accepted Submission(s): 15293 Problem Description 某国为了防御敌国的导弹袭击,发展出一种导弹拦截系统.但是这种导弹拦截系统有一个缺陷:虽然它的第一发炮弹能够到达任意的高度,但是以后每一发炮弹都不能超过前一发的高…

Tickets Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4393    Accepted Submission(s): 2236 Problem Description Jesus, what a great movie! Thousands of people are rushing to the cinema. However…

尽量沿着边走距离最短.化减后 C(n+1,k)+ n - k, 预处理阶乘,Lucas定理组合数取模 DP? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 128000/128000 K (Java/Others) Total Submission(s): 1899    Accepted Submission(s): 633 Problem Description Figure 1 shows the Yang Hui Tri…

简单的二分图匹配: 每个位置可以边到这些数字甚至可以边 Game with Pearls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 122    Accepted Submission(s): 85 Problem Description Tom and Jerry are playing a game with tubes an…

一个模式的dp. Pearls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1096    Accepted Submission(s): 476 Problem Description In Pearlania everybody is fond of pearls. One company, called The Royal Pe…

杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4472 Problem Description Prof. Tigris is the head of an archaeological team who is currently in charge of an excavation in a site of ancient relics. This site contains relics of a village where civilizat…

转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…

原题连接:http://acm.hdu.edu.cn/showproblem.php?pid=1300 题意: 题目太长了..自己看吧 题解: 看懂题目,就会发现这是个傻逼dp题,斜率优化一下就好 代码: #include<iostream> #include<cstring> #include<vector> #include<cstdio> #define MAX_N 500 using namespace std; typedef long long…

http://acm.hdu.edu.cn/showproblem.php?pid=5009 2014网络赛 西安 比较难的题 Paint Pearls Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1951    Accepted Submission(s): 631 Problem Description Lee has a str…

Paint Pearls Problem Description   Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help. In e…

题目传送门 /* 题意:给n块砖头,问能组成多少个楼梯,楼梯至少两层,且每层至少一块砖头,层与层之间数目不能相等! 递推DP:dp[i][j] 表示总共i块砖头,最后一列的砖头数是j块的方案数 状态转移方程:dp[i][j] += dp[i-j][k] 表示最后一列是j,那么上一个状态是少了最后一列 总共i-j块砖头,倒数第二列是k块砖头.k<j, j<=i 最后累加dp[n][i], i<n因为最少要两层 dp[0][0] = 1; 还有更简单的做法,没看懂:http://m.blog…

题目传送门 /* 找规律/数位DP:我做的时候差一点做出来了,只是不知道最后的 is_one () http://www.cnblogs.com/crazyapple/p/3315436.html 数位DP:http://blog.csdn.net/libin56842/article/details/11580497 */ #include <cstdio> #include <iostream> #include <algorithm> #include <c…

http://poj.org/problem?id=1260 Pearls Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8474   Accepted: 4236 Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls…

题目传送门 /* 递推DP: 如果a, b, c是等差数列,且b, c, d是等差数列,那么a, b, c, d是等差数列,等比数列同理 判断ai-2, ai-1, ai是否是等差(比)数列,能在O(n)时间求出最长的长度 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long ll…

题目传送门 /* 题意:给一个区间,问任意两个数的素数因子的GCD最大 数学+dp:预处理出f[i],发现f[i] <= 7,那么用dp[i][j] 记录前i个f[]个数为j的数有几个, dp[r][j] - dp[l-1][j]表示区间内j的个数,情况不多,分类讨论一下 */ #include <cstdio> #include <algorithm> #include <cstring> #include <vector> #include <…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6647   Accepted: 3241 Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its name be…

Pearls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2018    Accepted Submission(s): 953 Problem Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, pro…

Problem Description Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help.  In each operation,…

// l表示从l[i]到i连续大于a[i]的最远左区间.r表示从i到r[i]连续大于a[i]的最远又区间 DP 找出 a[i] 的最远左区间和最远右区间与自己连着的比自己大的数的长度 , 然后用这个长度乘以 a[i], 乘积最大的那个就是答案 hdoj 1506 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define N 100000+10 #define…

题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1017 搞了一个下午才弄出来,,,,, 还是线性DP做的不够啊 看过数据量就知道状态转移方程和x,没有关系 然后再仔细推导就会知道整个转移方程只和k与y的差值有关 然后继续推导.... 定义dp[i][j]表示表示第j步删除最上边的i点, 定义mv[i]表示删除i点的同时可以向下移动的最远位置(可以预处理出来) 那么dp[i][j]=max(dp[i-1][j],dp[i-mv…

Pearls Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 10558   Accepted: 5489 Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its…

这道题几经波折啊. 最开始和vfleaking一样,把题意理解错了,认为一个装备可能被多个装备依赖,然后想不出来,去看题解. 发现自己理解错了题意,自己想想,其实也不难想到dp[i][j][k]表示“i号节点代表的子树,用掉j的钱,给父亲预留k个自己(但还是父亲付钱)”的状态,写出来交上去就是T, 开始以为是常数问题,优化半天还是T,看了他人AC代码,才发现自己算法有一定问题:重复计算了很多东西. 树形动规,一般在大的树规下,每个节点还会对儿子们来一次”资源分配“,一般是用背包解决,这道题也是这…

HDOJ(HDU).2844 Coins (DP 多重背包+二进制优化) 题意分析 先把每种硬币按照二进制拆分好,然后做01背包即可.需要注意的是本题只需要求解可以凑出几种金钱的价格,而不需要输出种数.因此用0表示不可以,1表示可以.最后对dp数组扫描一遍即可. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 100…

HDOJ(HDU).1059 Dividing(DP 多重背包+二进制优化) 题意分析 给出一系列的石头的数量,然后问石头能否被平分成为价值相等的2份.首先可以确定的是如果石头的价值总和为奇数的话,那么肯定不能被平分.若为偶数,则对valuesum/2为背包容量,全体石头为商品做完全背包.把完全背包进行二进制优化后,转为01背包即可. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #inclu…

HDOJ(HDU).2191. 悼念512汶川大地震遇难同胞――珍惜现在,感恩生活 (DP 多重背包+二进制优化) 题意分析 首先C表示测试数据的组数,然后给出经费的金额和大米的种类.接着是每袋大米的价格,重量和袋数. 每种大米的数量是有限的,应该能看出是多重背包的问题.关键是多重背包的处理方法.对多重背包采用二进制优化的方法.设同种大米的数量为n,每袋的价格为v,那么把同一种大米分别拆成1,2,4,8,16--(直到不够2^n时,剩下的单独分成一组),拆分的价格也分别对应是v,2v,4v,8v…

HDOJ(HDU).2159 FATE (DP 带个数限制的完全背包) 题意分析 与普通的完全背包大同小异,区别就在于多了一个个数限制,那么在普通的完全背包的基础上,增加一维,表示个数.同时for循环多写一层即可. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 105 using namespace std; int dp[…