zoukankan      html  css  js  c++  java
  • Codeforces687C(SummerTrainingDay03-D DP)

    C. The Values You Can Make

    time limit per test:2 seconds
    memory limit per test:256 megabytes
    input:standard input
    output:standard output

    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 xusing 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.

    Examples

    input

    6 18
    5 6 1 10 12 2

    output

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

    input

    3 50
    25 25 50

    output

    3
    0 25 50
     1 //2017-08-16
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <iostream>
     5 #include <algorithm>
     6 
     7 using namespace std;
     8 
     9 const int N = 510;
    10 //dp[i][j]表示选取某些数使其和为i的时候,能否组成j。转移方程:dp[i][j] = dp[i][j+arr[p]] = 1 | dp[i-arr[p]][j] == 1
    11 int arr[N], dp[N][N], ans[N], tot;
    12 int n, k;
    13 bool vis[N], book[N];
    14 
    15 int main()
    16 {
    17     while(cin>>n>>k){
    18         for(int i = 0; i < n; i++){
    19               cin>>arr[i];
    20         }
    21         memset(dp, 0, sizeof(dp));
    22         dp[0][0] = 1;
    23         for(int p =  0; p < n; p++){
    24             for(int i = k; i >= arr[p]; i--){
    25                 for(int j = 0; j+arr[p] <= k; j++)
    26                     if(dp[i-arr[p]][j]){
    27                         dp[i][j] = dp[i][j+arr[p]] = 1;
    28                     }
    29             }
    30         }
    31         tot = 0;
    32         for(int j = 0; j <= k; j++){
    33             if(dp[k][j])ans[tot++] = j;
    34         }
    35         cout<<tot<<endl;
    36         for(int i = 0; i < tot; i++)
    37               if(i == tot-1)cout<<ans[i]<<endl;
    38             else cout<<ans[i]<<" ";
    39     }
    40 
    41     return 0;
    42 }
  • 相关阅读:
    Spring 事务传播实践分析
    记一次%转义引发的血案
    Springboot+redis 整合
    SpringBoot基础梳理
    MyBatis String类型传递参数注意事项
    SpringBoot填坑系列---XML方式配置数据库
    自定义AlertView(Swift)
    iOS开发,最新判断是否是手机号的正则表达式
    iOS开发 UILabel实现自适应高宽
    iOS开发笔记--UILabel的相关属性设置
  • 原文地址:https://www.cnblogs.com/Penn000/p/7374074.html
Copyright © 2011-2022 走看看