Cut Ribbon
Polycarpus has a ribbon, its length is n. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length a, b or c.
- After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
The first line contains four space-separated integers n, a, b and c (1 ≤ n, a, b, c ≤ 4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers a, b and c can coincide.
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
5 5 3 2
2
7 5 5 2
2
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
#define ll long long
#define mod 998244353
using namespace std;
int dir[][] = { {,},{,-},{-,},{,} }; int main()
{
int n;
cin >> n;
vector<int> a(),dp(n+);
for (int i = ; i < ; i++)
cin >> a[i];
sort(a.begin(), a.end());
if (a[] <= n) dp[a[]] = ;
if (a[] <= n) dp[a[]] = ;
if (a[] <= n) dp[a[]] = ;
for (int i = ; i <= n; i++)
{
int ans = dp[i];
if (i > a[] && dp[i - a[]] != ) ans = max(ans, dp[i - a[]] + );
if (i > a[] && dp[i - a[]] != ) ans = max(ans, dp[i - a[]] + );
if (i > a[] && dp[i - a[]] != ) ans = max(ans, dp[i - a[]] + );
dp[i] = ans;
}
cout << dp[n] << endl;
return ;
}
Cut Ribbon的更多相关文章
- Codeforces Round #119 (Div. 2) Cut Ribbon(DP)
Cut Ribbon time limit per test 1 second memory limit per test 256 megabytes input standard input out ...
- Codeforces Round #119 (Div. 2)A. Cut Ribbon
A. Cut Ribbon time limit per test 1 second memory limit per test 256 megabytes input standard input ...
- Codeforces 189A:Cut Ribbon(完全背包,DP)
time limit per test : 1 second memory limit per test : 256 megabytes input : standard input output : ...
- [CF189A]Cut Ribbon(完全背包,DP)
题目链接:http://codeforces.com/problemset/problem/189/A 题意:给你长为n的绳子,每次只允许切a,b,c三种长度的段,问最多能切多少段.注意每一段都得是a ...
- 【CF 189A Cut Ribbon】dp
题目链接:http://codeforces.com/problemset/problem/189/A 题意:一个长度为n的纸带,允许切割若干次,每次切下的长度只能是{a, b, c}之一.问最多能切 ...
- CF 189A Cut Ribbon
#include<bits/stdc++.h> using namespace std; const int maxn = 4000 + 131; int n, a, b, c; int ...
- Codeforces 189A. Cut Ribbon
题目链接:http://codeforces.com/problemset/problem/189/A 题意: 给你一个长度为 N 的布条, 再给你三个长度 a, b , c.你可以用剪刀去剪这些布条 ...
- Codeforces Round #119 (Div. 2)
A. Cut Ribbon \(f(i)\)表示长为\(i\)的布条最多可以剪几段. B. Counting Rhombi \(O(wh)\)枚举中心计算 C. Permutations 将序列一映射 ...
- 使用vs2010创建MFC C++ Ribbon程序
Your First MFC C++ Ribbon Application with Visual Studio 2010 Earlier this month, I put together my ...
随机推荐
- cc.fade.fade
用cc.fadeIn之前要先把setOpacity(0),笑哭,啊啊啊啊啊.因为这个东西卡了好久,啊啊啊
- 在什么情况下,不写notify()或者notifyAll()就能唤醒被wait()阻塞的线程?
之前再看java关于线程的某视频时,发现在JDK源码中,join()=join(0)=wait()=wait(0),但是视频中在join()了之后,并没有用notify()或者notifyAll()去 ...
- ArcMap 导出Table数据到Excel
- 3、const与constexpr
初遇到constexpr真的是有点懵比,看了很多博客也没看懂,不知道是我太笨,还是别人写的太深奥?总之经过一番折腾算是入门了.一下是我个人总结,有不对的地方望指出. 一.学习const与constex ...
- The Softmax function and its derivative
https://eli.thegreenplace.net/2016/the-softmax-function-and-its-derivative/ Eli Bendersky's website ...
- PP: Sequence to sequence learning with neural networks
From google institution; 1. Before this, DNN cannot be used to map sequences to sequences. In this p ...
- Appium+Python移动端(Android)自动化测试环境搭建
一.安装JDK 下载好jdk安装包后直接下一步直至安装完成即可,安装完JDK后配置环境变量 :计算机→属性→高级系统设置→高级→环境变量: 系统变量→新建 JAVA_HOME 变量 变量值填写jdk的 ...
- VM中Linux网络设置(固定ip、连接外网开发环境)
在开发过程中,我们经常需要在linux中进行操作.毕竟服务器的系统大多数都是Linux,所以在dev环境需要配置好一台Linux系统配合开发. 在VMWare Workstation Pro中 ...
- java的服务是每收到一个请求就新开一个线程来处理吗?tomcat呢?
首先,服务器的实现不止有这两种方式. 先谈谈题主说的这两种服务器模型: 1.收到一个请求就处理,这个时候就不能处理新的请求,这种为阻塞 这个是单线程模型,无法并发,一个请求没处理完服务器就会阻塞,不会 ...
- JS的冒泡事件
在一个对象上触发某类事件(比如单击onclick事件),如果此对象定义了此事件的处理程序,那么此事件就会调用这个处理程序,如果没有定义此事件处理程序或者事件返回true,那么这个事件会向这个对象的 ...