zoukankan      html  css  js  c++  java
  • Codeforces Round #360 (Div. 2) E. The Values You Can Make dp

    E. The Values You Can Make

    题目连接:

    http://www.codeforces.com/contest/688/problem/E

    Description

    Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin is ci. The price of the chocolate is k, so Pari will take a subset of her coins with sum equal to k and give it to Arya.

    Looking at her coins, a question came to her mind: after giving the coins to Arya, what values does Arya can make with them? She is jealous and she doesn't want Arya to make a lot of values. So she wants to know all the values x, such that Arya will be able to make x using some subset of coins with the sum k.

    Formally, Pari wants to know the values x such that there exists a subset of coins with the sum k such that some subset of this subset has the sum x, i.e. there is exists some way to pay for the chocolate, such that Arya will be able to make the sum x using these coins.

    Input

    The first line contains two integers n and k (1  ≤  n, k  ≤  500) — the number of coins and the price of the chocolate, respectively.

    Next line will contain n integers c1, c2, ..., cn (1 ≤ ci ≤ 500) — the values of Pari's coins.

    It's guaranteed that one can make value k using these coins.

    Output

    First line of the output must contain a single integer q— the number of suitable values x. Then print q integers in ascending order — the values that Arya can make for some subset of coins of Pari that pays for the chocolate.

    Sample Input

    6 18
    5 6 1 10 12 2

    Sample Output

    16
    0 1 2 3 5 6 7 8 10 11 12 13 15 16 17 18

    Hint

    题意

    所有能组成K的C的子集方案中,能拼出哪些面额

    题解

    n^3dp

    dp[i][j][k]表示用到了第i个数,当前和为j,子集和为k可不可行

    裸的会稍微卡一下空间,这个滚动数组优化,或者直接用bool就可以了。

    代码

    #include<bits/stdc++.h>
    using namespace std;
    const int maxn = 505;
    
    int n,K;
    bool dp[maxn][maxn][maxn];
    int main(){
        scanf("%d%d",&n,&K);
        dp[0][0][0]=1;
        for(int i=1;i<=n;i++){
            int c;
            scanf("%d",&c);
            for(int j=0;j<=500;j++){
                for(int k=0;k<=j;k++){
                    dp[i][j][k]|=dp[i-1][j][k];
                    if(j>=c)dp[i][j][k]|=dp[i-1][j-c][k];
                    if(k>=c)dp[i][j][k]|=dp[i-1][j-c][k-c];
                }
            }
        }
        vector<int> ans;
        for(int i=0;i<=500;i++)
            if(dp[n][K][i])ans.push_back(i);
        cout<<ans.size()<<endl;
        for(int i=0;i<ans.size();i++)
            cout<<ans[i]<<" ";
        cout<<endl;
    }
  • 相关阅读:
    bash快捷建-光标移到行首、行尾等
    VitualBox中linux系统ping ip能通域名不通的解决办法
    linux下使用tar命令
    Windows平台下Git服务器搭建
    android 在使用studio 编写百度地图中遇到APP Scode码校验失败 问题
    android 开发中 添加库文件 和so 文件的存放位置和添加依赖
    Volley之 JsonRequest 解析JSON 数据
    利用Volley封装好的图片缓存处理加载图片
    使用Volley执行网络数据传输
    android 测试 Monkey 和 MonkeyRunner 的使用
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5631208.html
Copyright © 2011-2022 走看看