Thief in a Shop n个物品每个价值ai,要求选k个,可以重复.问能取到哪几个价值? 1 ≤ n, k ≤ 1000,1 ≤ ai ≤ 1000 题解 将选一个物品能取到的价值的01生成函数k次方即可得到选k个物品得到的某个权值的方案数. 出题人卡NTT模数,998244353和1004535809都会被卡.然而469762049没被卡-- CO int N=1048576; int a[N]; int rev[N],omg[N]; void NTT(int a[],int lim)…

题目大意:有一个小偷,拿$k$个东西,有$n$种产品,每种产品都有无限多个.对于每个第$i$ 种产品,它的价值是$A_i$.可能偷走的物品价值之和. 题解:对于所有的物品构造生成函数$F(x)=\sum\limits_{i\in A}x^i$,取$k$个物品相当于取其中的$k$项相乘,输出$F^k(x)$中不为零的项就行了.(这道题模数$998244353$和$1004535809$都被$hack$了,看$Weng\_weijie\;dalao$的题解得双模数没被卡,于是就$A$了)(这道题似乎…

E. Thief in a Shop time limit per test 5 seconds memory limit per test 512 megabytes input standard input output standard output A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. There ar…

题目链接 E. Thief in a Shop time limit per test 5 seconds memory limit per test 512 megabytes input standard input output standard output A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. The…

C - Thief in a Shop 思路 :严格的控制好k的这个数量,这就是个裸完全背包问题.(复杂度最极端会到1e9) 他们随意原来随意组合的方案,与他们都减去 最小的 一个 a[ i ] 组合的方案数目是不会改变的 那么我们就 dp [ i ]表示 i 这个价格需要的最少 个数.  这样求最小个数保证不会漏解 然后 如果这个  i 能通过 1 - k 个物品组合出来,那么 一定能通过k 个物品组合出 i + k * a [ 1 ]. #include<bits/stdc++.h> us…

E - Thief in a Shop 题目大意:给你n ( n <= 1000)个物品每个物品的价值为ai (ai <= 1000),你只能恰好取k个物品,问你能组成哪些价值. 思路:我们很容易能够想到dp[ i ][ j ]表示取i次j是否存在,但是复杂度1e12肯定不行. 我们将ai排序,每个值都减去a[1]然后再用dp[ i ]表示到达i这个值最少需要取几次,只需要1e9就能完成, 我们扫一遍dp数组,如果dp[ i ]  <= k 则说明 i + k * a[1]是能取到的.…

E. Thief in a Shop 题目连接: http://www.codeforces.com/contest/632/problem/E Description A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an inf…

E. Thief in a Shop   A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product o…

A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product of kind i is ai. The t…

题目 Source http://codeforces.com/contest/632/problem/E Description A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of pro…

A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product of kind i is ai. The t…

题意:n个物品每个价值a[i],要求选k个,可以重复,问能取到哪几个价值 题解:fft裸题.但是直接一次fft,然后快速幂会boom.这样是严格的\(2^{20}*log2(2^{20})*log(w)\).需要在快速幂里fft,每次取最大的2的次幂,然后fft也boom了,不知道是不是写搓了.ntt过了..... //#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4) //#pragma GCC…

传送门 题意简述:给nnn个物件,物件iii有一个权值aia_iai​,可以选任意多个.现在要求选出kkk个物件出来(允许重复)问最后得到的权值和的种类数. n,k,ai≤1000n,k,a_i\le1000n,k,ai​≤1000 思路: 这是一道很显然的生成函数,我们把选一个物件的生成函数给列出来,然后取它的kkk次方就是答案. 显然可以上一波fftfftfft 成功T飞 在博主卡场无果之后换成了nttnttntt,然后发现cfcfcf的强力数据同时卡掉了998244353998244353…

题意:有N种物品,每种物品有价值\(a_i\),每种物品可选任意多个,求拿k件物品,可能损失的价值分别为多少. 分析:相当于求\((a_1+a_2+...+a_n)^k\)中,有哪些项的系数不为0.做k次FFT求卷积求卷积肯定爆炸,考虑用分治的形式计算,因为中间计算的时候会重复计算一些幂次,所以用记忆化搜索的形式,保留计算结果. 因为只要计算出哪些项不为0,所以卷积之后求结果时,系数非0项用1作系数即可,否则分分钟炸精度. 当然也可以用快速幂求解#. #include <bits/stdc++.…

题意:给你n种物品,每种无限个,问恰好取k个物品能组成哪些重量.n<=1000,k<=1000,每种物品的重量<=1000. 我们搞出选取一种物品时的生成函数,那么只要对这个生成函数求k次幂就可以了.结果会很大所以我们可以在模意义下NTT来搞.然而会有一个问题,就是算出来的系数可能恰好是模数的倍数,比如说我们只用998244353就会WrongAnswer on test 20.然后我试了和异化多肽的模数1005060097一块来,然后,WrongAnswer on test 5…Exc…

题意: 问n个物品选出K个可以拼成的体积有哪些. 解法: 多项式裸题,注意到本题中 $A(x)^K$ 的系数会非常大,采用NTT优于FFT. NTT 采用两个 $2^t+1$ 质数,求原根 $g_n$ 后用 $g_n^1 $~$ g_n^{P-1}$ 的循环代替复数向量的旋转. 注意逆的 $w_n$ 是 $g_n ^ {  - \frac{P-1}{len}  }$,并且要用两个质数保证正确即可,$O(nlogn)$. #include <bits/stdc++.h> #define PI a…

[链接] 我是链接,点我呀:) [题意] n个物品,每个物品都有无限个. 第i个物品的价格是一样都,都是ai 让你从中选出恰好k个物品 问你选出的物品的总价值 有多少种不同的可能. [题解] 可以用f[j]表示物品的总价值为j最少需要选多少个物品. for (int i = 1;i <= n;i++) for (int j = a[i];j <= k*a[i];j++){ f[j] = min(f[j],f[j-a[i]]+1); } 然后对于价值j; 如果f[j]==k的话.j就可行? 但是…

632E:http://codeforces.com/problemset/problem/632/E 参考:https://blog.csdn.net/qq_21057881/article/details/51023067 题意: 给定n个值,让你选择k个数,可以重复选择,问可以得到哪些数字. 思路: 显然最小的值起到很大的作用,我们可以把每个值都减去这个最小值,利用完全背包,建立dp[i]表示,取到 i 这么多值最少需要多少个数.如果取到 i值需要的数值小于等于K,那么k * 最小值 +…

众所周知,tzc 在 2019 年(12 月 31 日)就第一次开始接触多项式相关算法,可到 2021 年(1 月 1 日)才开始写这篇 blog. 感觉自己开了个大坑( 多项式 多项式乘法 好吧这个应该是多项式各种运算中的基础了. 首先,在学习多项式乘法之前,你需要学会: 复数 我们定义虚数单位 \(i\) 为满足 \(x^2=-1\) 的 \(x\). 那么所有的复数都可以表示为 \(z=a+bi\) 的形式,其中 \(a,b\) 均为实数. 复数的加减直接对实部虚部相加减就行了. 复数的乘…

Educational Codeforces Round 9 Longest Subsequence 题目描述:给出一个序列,从中抽出若干个数,使它们的公倍数小于等于\(m\),问最多能抽出多少个数,并输出方案与公倍数. solution 枚举每一个数,将它的倍数的答案加一,最大值就是答案. 时间复杂度:\(O(nlogn)\) Thief in a Shop 题目描述:给出\(n\)中物品与它们的价值,每种物品都有无限个,问从中拿出\(m\)个,价值有可能是什么,输出所有的价值. soluti…

数学基础(卷积,FFT,FWT,FMT,鸽巢原理,群论,哈里亚余数,哈里亚计数定理,组合数学,LVG定理,期望DP,期望点贡献问题) 练习题: A - Necklace of Beads Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are produced by rotation a…

注:多项式的题目,数组应开:N的最近2的整数次幂的4倍. 多项式乘法 FFT模板 时间复杂度\(O(n\log n)\). 模板: void FFT(Z *a,int x,int K){ static int rev[N],lst; int n=(1<<x); if(n!=lst){ for(int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<x-1); lst=n; } for(int i=0;i<…

Description You want to arrange the window of your flower shop in a most pleasant way. You have F bunches of flowers, each being of a different kind, and at least as many vases ordered in a row. The vases are glued onto the shelf and are numbered con…

Magicodes.Shop为湖南心莱信息科技有限公司(xin-lai.com)Magicodes系列产品之一. 产品中引用的Magicodes系列Nuget包的开源库地址为:https://github.com/xin-lai/ 购买地址:https://item.taobao.com/item.htm?id=539695057814 V1.2 (2016-11-19) 修改下单后,没有获取用户手机号,送货地址的信息等 后台商品类目管理,隐藏logo的显示 修复个人中心底部链接问题 新增商品评…

104. Little shop of flowers time limit per test: 0.25 sec. memory limit per test: 4096 KB 问题: 你想要将你的花窗安排得最具美感.有F束花,每一束花都不一样,至少有F个按顺序排成一行的花瓶.花瓶从左到右,依次编号1-V.而花放置的位置是可以改变的,花依次编号1到F.花的序号有一个特征,即是序号决定了花束出现在花瓶里的顺序.例如,有两束花编号i和j,满足i<j,那么i所在的花瓶一定要在j所在花瓶的左边,即是i…

第一节 初始目录结构 (1)初识目录结构 在创建应用之前,我们来看一下Yii 1.x版本的目录结构:将yii-1.1.13安装文件解压到网站根目录下,打开framework目录,其目录如下图所示 (2)认识framework目录 第二节 命令行创建应用~shop (1)创建应用 进入framework目录,通过如下命令行创建一个商场系统(shop) 通过如上命令创建一个商场系统(shop)后,yiitest目录下自动生成shop目录 (2)访问shop 通过如下地址即可访问刚创建的商城系统(sh…

[POJ1157]LITTLE SHOP OF FLOWERS 试题描述 You want to arrange the window of your flower shop in a most pleasant way. You have F bunches of flowers, each being of a different kind, and at least as many vases ordered in a row. The vases are glued onto the s…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20802   Accepted: 9613 Description You want to arrange the window of your flower shop in a most pleasant way. You have F bunches of flowers, each being of a different kind, and at least as…

TIANKENG's rice shop 题目链接: http://acm.hust.edu.cn/vjudge/contest/123316#problem/J Description TIANKENG managers a pan fried rice shop. There are n kinds of fried rice numbered 1-n. TIANKENG will spend t time for once frying. Because the pan is so sma…

Alice's mooncake shop Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4122 Description The Mid-Autumn Festival, also known as the Moon Festival or Zhongqiu Festival is a popular harvest festival celebrated by Ch…