[AtCoder] C - Tak and Cards 传送门:https://atcoder.jp/contests/abc044/tasks/arc060_a 挺容易想到的dp,但是因为脑残改了一晚上,最后出错的地方自己都看笑了 //数据很小,复杂度最高但最好理解的三维dp也能过 1#include<cstdio>2#include<cstring>3#include<cstdlib>4#include<iostream>5#include<algo...
C - Tak and Cards (atcoder.jp) 题意: 给定一个数组 A , 以及一个数字 x , 如果要从 A 中选出若干个数,使得平均数为x,有多少种选择? 分析: 如果要纯讨论的话,感觉无从下手,我们直接考虑dp f[i][j][k] 我们如果用来表示前i个数中,选择了j个数,此时的平均数为k,可以发现状态转移的时候不怎么...
AtCoder-arc060 (题解) A - 高橋君とカード / Tak and Cards (DP) 题目链接 题目大意: 有nn个数字,要求取出一些数字,使得它们的平均数恰好为xx,问有几种取法。 大致思路: 只要将每一个数字减掉xx,那么问题就变成在nn个数字中选取一些数字使得和为00的方案数,是一个经典的dpdp问题,不过要注意细节问题 ...
C_AtCoder_Cards.cpp C_AtCoder_Magics.cpp C_Avoid_K_Palindrome_2.cpp C_Baby_Ehab_Partitions_Again.cpp C_Bad_Sequence.cpp C_Bad_Sequence.exe C_Bakry_and_Partitioning.cpp C_Balanced_Stone_Heaps.cpp C_Ball_in_Berland.cpp C_Balls_and_Bag_Query.cpp C_Banknotes.cpp C_Basil_s_Garden.cpp...
AtCoder Regular Contest 078-C - Splitting Pile C - Splitting Pile Time limit : 2sec / Memory limit : 256MB Score : 300 points Problem Statement Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer ai written on it. They will share these cards. ...
The i-th card from the top has the integer ai written on it. They will share these cards. First, Snuke will take some number of ...AtCoder Beginner Contest 050 C - Lining Up C - Lining Up 题目大意: 有n个人 编号是从1到n,他们不知道自己所在的位置,但是他们知道自己左边和右边的人数...
题目你有N根竹竿子,这N根竹竿子的长度依次为l1,l2,l3..lN(单位是厘米)你的任务是,从N根竹竿子里面选择一些竹竿子(全部选择也可以),用这些选出来的竹竿子...
题目链接:http://abc044.contest.atcoder.jp/tasks/arc060_a Time limit : 2sec / Memory limit : 256MB Score :300points Problem Statement Tak hasNcards. On thei-th(1≤i≤N)card is written an integerxi. He is selecting one or more cards from theseNcards, so that the average of the ...
https://atcoder.jp/contests/abc213/tasks/abc213_c眺めてて思ったが、行列とか関係なくね?それぞれの要素(A1,B1...An,Bn)を小さい順に …
// Problem: C - Tak and Cards // URL: https://atcoder.jp/contests/abc044/tasks/arc060_a // Author: Pengfei Xu #include<bits/stdc++.h> using namespace std; using ll = long long; const int maxn = 55; int a[maxn]; ll f[maxn*maxn]; void solve(){ int n,x; cin >> n...