HDOJ(HDU).2660 Accepted Necklace (DFS) 点我挑战题目 题意分析 给出一些石头,这些石头都有自身的价值和重量.现在要求从这些石头中选K个石头,求出重量不超过W的这些石头的最大价值是多少? 类似于之前讨论到的数字选不选的问题,此处面临的情况是石头选不选,若选进行一个dfs,若不选择进行另外一个dfs.考虑递归边界: 1.当选够了K个的时候,终止递归: 2.当当前重量大于W的时候,终止递归: 3.当所选石头的下标(代码中的pos)超过石头数量的时候,终止递归: 若…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2660 Accepted Necklace Description I have N precious stones, and plan to use K of them to make a necklace for my mother, but she won't accept a necklace which is too heavy. Given the value and the weight…

Accepted Necklace Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2474    Accepted Submission(s): 973 Problem Description I have N precious stones, and plan to use K of them to make a necklace f…

Problem Description I have N precious stones, and plan to use K of them to make a necklace for my mother, but she won't accept a necklace which is too heavy. Given the value and the weight of each precious stone, please help me find out the most valu…

http://acm.hdu.edu.cn/showproblem.php?pid=2660 f[v][u]=max(f[v][u],f[v-1][u-w[i]]+v[i]; 注意中间一层必须逆序循环. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ][]; ],w[]; int main() { //freopen("a.txt",&qu…

Accepted Necklace Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3136 Accepted Submission(s): 1213 Problem Description I have N precious stones, and plan to use K of them to make a necklace for m…

hdu 5730 Shell Necklace 题意:求递推式\(f_n = \sum_{i=1}^n a_i f_{n-i}\),模313 多么优秀的模板题 可以用分治fft,也可以多项式求逆 分治fft 注意过程中把r-l+1当做次数界就可以了,因为其中一个向量是[l,mid],我们只需要[mid+1,r]的结果. 多项式求逆 变成了 \[ A(x) = \frac{f_0}{1-B(x)} \] 的形式 要用拆系数fft,直接把之前的代码复制上就可以啦 #include <iostream…

Problem Description I have N precious stones, and plan to use K of them to make a necklace for my mother, but she won't accept a necklace which is too heavy. Given the value and the weight of each precious stone, please help me find out the most valu…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5730 [题目大意] 给出一个数组w,表示不同长度的字段的权值,比如w[3]=5表示如果字段长度为3,则其权值为5,现在有长度为n的字段,求通过不同拆分得到的字段权值乘积和. [题解] 记DP[i]表示长度为i时候的答案,DP[i]=sum_{j=0}^{i-1}DP[j]w[i-j],发现是一个卷积的式子,因此运算过程可以用FFT优化,但是由于在计算过程中DP[j]是未知值,顺次计算复杂度是O(…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=5730 可以用分治FFT.但自己只写了多项式求逆. 和COGS2259几乎很像.设A(x),指数是长度,系数是方案. \( A(x)^{k} \) 的 m 次项系数表示 k 个连续段组成长度为 m 的序列的方案数. \( B(x)=1+F(x)+F^{2}(x)+F^{3}(x)+... \) \( B(x) = \frac{1}{1-F(x)} \)(通过计算B(x)的逆来看出这个式子) 然后多项式求逆…