CRB and His Birthday Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1112    Accepted Submission(s): 559 Problem Description Today is CRB's birthday. His mom decided to buy many presents for her…

题目地址:HDU 5410 题意:有M元钱,N种礼物,若第i种礼物买x件的话.会有Ai*x+Bi颗糖果,现给出每种礼物的单位价格.Ai值与Bi值.问最多能拿到多少颗糖果. 思路:全然背包问题. dp[j][1]在当前物品时花钱为j的而且买过当前物品的最大值. dp[j][0]不买当前这件物品此前花钱为j的的最大值. 每种物品的价值随Ai线性变化,可是不随B[i]线性变化.B[i]仅是在第一次挑选第i件物品是才算入,其它时候均不算入. 所以我们能够写出状态转移方程: dp[j][0]=max(dp…

CRB and His Birthday Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 430    Accepted Submission(s): 237 Problem Description Today is CRB's birthday. His mom decided to buy many presents for her…

Problem Description Today is CRB's birthday. His mom decided to buy many presents for her lovely son. She went to the nearest shop with M Won(currency unit). At the shop, there are N kinds of presents. It costs Wi Won to buy one present of i-th kind.…

对于每个物品,如果购买,价值为A[i]*x+B[i]的背包问题. 先写了一发是WA的= =.代码如下: #include <stdio.h> #include <algorithm> #include <string.h> #include <set> using namespace std; typedef pair<int,int> pii; pii dp[]; ],A[],B[]; int main() { int T;scanf(&quo…

题目大意: 一个人要去买礼物,有M元.有N种礼物,每件礼物的价值是Wi, 你第i件礼物买k个 是可以得到 Ai * k + Bi 个糖果的. 问怎么才能使得你得到的糖果数目最多.   其实就是完全背包了.物品的个数是有多个的. dp[第n件物品][已经花费了m元]   DP式子:    dp[n][m] = max(dp[n-1][m-W[n]] + A[i]+B[i], dp[n][m-W[i]] + A[i]); 最后结束的时候把上面的结果挪下来            for(i=0; i<…

题意:有n种商品,每种商品中有a个糖果,如果买这种商品就送多b个糖果,只有第一次买的时候才送.现在有m元,最多能买多少糖果? 思路:第一次买一种商品时有送糖果,对这一次进行一次01背包,也就是只能买一次.然后对这种商品来一次完全背包,此时不送糖果,也可以多买. #include <bits/stdc++.h> #define pii pair<int,int> #define INF 0x7f7f7f7f #define LL long long using namespace s…

#include<stdio.h> #include<string.h> #include<vector> #include<queue> #include<algorithm> using namespace std; int main() { int _,i,j,m,n,k,a[1024],b[1024],w[1024],dp[2048]; scanf("%d\n",&_); while(_--) { scanf(…

CRB and Tree Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2112    Accepted Submission(s): 635 Problem Description CRB has a tree, whose vertices are labeled by 1, 2, …, N. They are connected…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5412 CRB and Queries Description There are $N$ boys in CodeLand. Boy i has his coding skill $A_{i}$. CRB wants to know who has the suitable coding skill. So you should treat the following two types of qu…

题目地址:pid=5407">HDU 5407 题意:CRB有n颗不同的糖果,如今他要吃掉k颗(0<=k<=n),问k取0~n的方案数的最小公倍数是多少. 思路:首先做这道题我们须要必备的几个技能点. 1. LCM(C(n,0), C(n,1),-, C(n,n))=LCM(1,2,3,-n+1)/(n+1).额,这个有一篇证明Kummer定理 2.(1) 乘法逆元定义: 满足a*k≡1 (mod p)的k值就是a关于p的乘法逆元(a,p互质). (2)为什么要用乘法逆元: 当…

题目链接: Hdu 5416 CRB and Tree 题目描述: 给一棵树有n个节点,树上的每条边都有一个权值.f(u,v)代表从u到v路径上所有边权的异或值,问满足f(u,v)==m的(u, v)有多少中情况(u, v有可能相同)? 解题思路: 由于xor的特殊性质.x^x=0,对于求f(u, v) == f(u, 1) ^ f(1, u). 又因为x^y == z可以推出x^z == y,对于f(u, 1) ^ f(1, v) == m可以转化为m ^ f(1, v) == f(u, 1)…

题目链接: Hdu 5407 CRB and Candies 题目描述: 给出一个数n,求lcm(C(n,0),C[n,1],C[n-2]......C[n][n-2],C[n][n-1],C[n][n])%(1e9+7)是多少? 解题思路: 刚开始的时候各种开脑洞,然后卡题卡的风生水起.最后就上了数列查询这个神奇的网站,竟然被我找到了!!!!就是把题目上给的问题转化为求lcm(1, 2, 3, 4 ...... n-2, n-1, n, n-1) / (n+1),扎扎就打了两个表一个lcm[n…

题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=5412 Problem Description There are N boys in CodeLand. Boy i has his coding skill Ai. CRB wants to know who has the suitable coding skill. So you should treat the following two types of queries. Query 1…

Problem Description There are N boys in CodeLand.Boy i has his coding skill Ai.CRB wants to know who has the suitable coding skill.So you should treat the following two types of queries.Query 1: 1 l vThe coding skill of Boy l has changed to v.Query 2…

CRB and Roads Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 495    Accepted Submission(s): 225 Problem Description There are N cities in Codeland.The President of Codeland has a plan to con…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=5411 题意:按题目转化的意思是,给定N和M,再给出一些边(u,v)表示u和v是连通的,问走0,1,2.....M步的方案数. 分析:这题和 hdu5318 The Goddess Of The Moon差点儿相同,就是多了一个等比数列求和. 代码: #include <cstdio> #include <iostream> #include <cstring> using na…

CRB and Candies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 722    Accepted Submission(s): 361 Problem Description CRB has N different candies. He is going to eat K candies.He wonders how ma…

CRB and Queries Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1602    Accepted Submission(s): 409 Problem Description There are N boys in CodeLand.Boy i has his coding skill Ai.CRB wants to…

CRB and Tree Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 967    Accepted Submission(s): 308 Problem Description CRB has a tree, whose vertices are labeled by 1, 2, …, N. They are connected b…

CRB and Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 483    Accepted Submission(s): 198 Problem Description CRB is now playing Jigsaw Puzzle. There are  kinds of pieces with infinite…

CRB and Tree                                                             Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)                                                                                            To…

CRB and Apple Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 421    Accepted Submission(s): 131 Problem Description In Codeland there are many apple trees.One day CRB and his girlfriend decide…

CRB and Candies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 453    Accepted Submission(s): 222 Problem Description   CRB has N different candies. He is going to eat K candies.He wonders how…

CRB and Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 301    Accepted Submission(s): 127 Problem DescriptionCRB is now playing Jigsaw Puzzle.There are N kinds of pieces with infinite su…

CRB and Queries Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 533    Accepted Submission(s): 125 Problem DescriptionThere are N boys in CodeLand.Boy i has his coding skill Ai.CRB wants to k…

欢迎參加--每周六晚的BestCoder(有米!) CRB and Tree Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 481    Accepted Submission(s): 151 Problem Description CRB has a tree, whose vertices are labeled by 1, 2,…

CRB and Queries Time Limit: 6000 MS Memory Limit: 131072 K Problem Description There are N boys in CodeLand. Boy i has his coding skill Ai. CRB wants to know who has the suitable coding skill. So you should treat the following two types of queries. Q…

CRB and Candies Problem's Link Mean: 给定一个数n,求LCM(C(n,0),C(n,1),C(n,2)...C(n,n))的值,(n<=1e6). analyse: 很有趣的一道数论题! 看了下网上别人的做法,什么Kummer定理我还真没听说过,仔细研究一下那个鬼定理真是涨姿势了! 然而这题我并不是用Kummer那货搞的(what?). 其实这题真的很简单(不要打我),为什么这样说呢?看了下面的解释你就知道我没骗你. 首先我们看一下这个式子:LCM(C(n,0…

题意:给一个正整数k,求lcm((k, 0), (k, 1), ..., (k, k)) 解法:在oeis上查了这个序列,得知答案即为lcm(1, 2, ..., k + 1) / (k + 1),而分子有一个递推式,如果k为一个质数x的某次幂,那么ans[k]为ans[k - 1] * m,否则ans[k] = ans[k - 1].做除法的时候用了逆元,因为取模来着. 代码: #include<stdio.h> #include<iostream> #include<al…