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
- Cenos7下指定ftp用户限制在特定目录下(亲身实践)
好了,废话不多说.上头下来个需求,让我给别人开个ftp账户,只能访问项目的目录,不能访问项目外的目录,就算cd切换目录也不行. 开始: 第一步;安装ftp,我用的是centos7,只需敲入命令 yum ...
- Excel时间格合并(年月日+时间点)
=value(a1)+b2 日期 时间 合并 2018/8/8 14:13 2018/8/8 14:13:00
- sqli-labs11-17(手注+sqlmap)
这几关涉及到的都是post型data处注入,和get型的差别就是注入点的测试处不一样,方法都是一样的 0x01 sqli-labs less-11 1.手工 由于是post型注入,那么我们不能在url ...
- Linux的语系的修改
1. 显示目前所支持的语系 [root@test ~]# echo $LANG 2. 修改语系成为英文语系 字体集目录:"/etc/sysconfig/i18n",编辑该文件,将字 ...
- Qt Gui 第五章绘图类
双缓冲 void Plotter::refreshPixmap() { pixmap = QPixmap(size()); pixmap.fill(, ); QPainter painter(& ...
- Normalizing flows
probability VS likelihood: https://zhuanlan.zhihu.com/p/25768606 http://sdsy888.me/%E9%9A%8F%E7%AC%9 ...
- Jmeter-第三方插件安装
1.插件下载 官方地址https://jmeter-plugins.org/install/Install/ 网盘地址链接: https://pan.baidu.com/s/1E4lnMWDGPWCN ...
- double加减乘除
//四舍五入 public static double toDecimal(Double num){ if(Double.isNaN(num) || num == null){ return 0; } ...
- TCP/IP协议和socket
1.传输层基于tcp协议的三次握手和四次挥手? 传输层有两种数据传输协议,分别为TCP协议和UDP协议,其中TCP协议为可靠传输,数据包没有长度设置,理论可以无限长,而UDP协议为不可靠传输,报头一共 ...
- 数据预处理 | 通过 Z-Score 方法判断异常值
判断异常值方法:Z-Score 计算公式 Z = (X-μ)/σ 其中μ为总体平均值,X-μ为离均差,σ表示标准差.z的绝对值表示在标准差范围内的原始分数与总体均值之间的距离.当原始分数低于平均值时, ...