Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she must pay the exact amount. Since she has as many as 1 coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find some coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (≤, the total number of coins) and M (≤, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the face values V​1​​≤V​2​​≤⋯≤V​k​​ such that V​1​​+V​2​​+⋯+V​k​​=M. All the numbers must be separated by a space, and there must be no extra space at the end of the line. If such a solution is not unique, output the smallest sequence. If there is no solution, output "No Solution" instead.

Note: sequence {A[1], A[2], ...} is said to be "smaller" than sequence {B[1], B[2], ...} if there exists k≥1 such that A[i]=B[i] for all i<k, and A[k] < B[k].

Sample Input 1:

8 9
5 9 8 7 2 3 4 1

Sample Output 1:

1 3 5

Sample Input 2:

4 8
7 2 4 3

Sample Output 2:

No Solution
 #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int N, M;
int coins[], dp[];
bool visit[][];
int main()
{
cin >> N >> M;
for (int i = ; i <= N; ++i)
cin >> coins[i];
sort(coins + , coins + N + , [](int a, int b) {return a > b; });
for (int i = ; i <= N; ++i)
{
for (int j = M; j >= coins[i]; --j)//目标递减
{
if (dp[j] <= dp[j - coins[i]] + coins[i])
{
dp[j] = dp[j - coins[i]] + coins[i];
visit[i][j] = true;//取
}
}
}
if (dp[M] != M)
cout << "No Solution" << endl;
else
{
vector<int>res;
int v = M, index = N;
while (v > )
{
if (visit[index][v] == true)
{
res.push_back(coins[index]);
v -= coins[index];
}
--index;
}
for (int i = ; i < res.size(); ++i)
cout << res[i] << (i == res.size() - ? "" : " ");
}
return ;
}

PAT甲级——A1068 Find More Coins的更多相关文章

  1. PAT甲级1068 Find More Coins【01背包】

    题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805402305150976 题意: n个硬币,每一个有一个特有的价 ...

  2. PAT 甲级 1068 Find More Coins

    https://pintia.cn/problem-sets/994805342720868352/problems/994805402305150976 Eva loves to collect c ...

  3. PAT 甲级 1068 Find More Coins(0,1背包)

    1068. Find More Coins (30) 时间限制 150 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Eva l ...

  4. PAT 甲级 1068 Find More Coins (30 分) (dp,01背包问题记录最佳选择方案)***

    1068 Find More Coins (30 分)   Eva loves to collect coins from all over the universe, including some ...

  5. 【PAT甲级】1048 Find Coins (25 分)(二分)

    题意: 输入两个正整数N和M(N<=10000,M<=1000),然后输入N个正整数(<=500),输出两个数字和恰好等于M的两个数(小的数字尽可能小且输出在前),如果没有输出&qu ...

  6. PAT甲级题解(慢慢刷中)

    博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给 ...

  7. 【转载】【PAT】PAT甲级题型分类整理

    最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel P ...

  8. PAT甲级题分类汇编——杂项

    本文为PAT甲级分类汇编系列文章. 集合.散列.数学.算法,这几类的题目都比较少,放到一起讲. 题号 标题 分数 大意 类型 1063 Set Similarity 25 集合相似度 集合 1067 ...

  9. PAT甲级题分类汇编——序言

    今天开个坑,分类整理PAT甲级题目(https://pintia.cn/problem-sets/994805342720868352/problems/type/7)中1051~1100部分.语言是 ...

随机推荐

  1. 007-Java可变个数形参重载【数组和...】

    重载方法时,可变个数形参的方法有两种方式 数组重载 ...重载 对两种方法,其实是一致的,示例如下: public class MethodArgsTest { //可变个数形参的格式:数据类型... ...

  2. IOS中iframe的滚动条不启作用

    引自:https://www.cnblogs.com/weinan/archive/2013/01/05/2832800.html 问题描述: iframe设置了高度(例如500px).倘若ifram ...

  3. java排序及泛型

    一.用泛型实现快排,可以传入不通类型进行排序,比如String数组,Integer数组. /** * 快速排序 * * @author chx * */ public class QuickSort ...

  4. 小白 Linux下安装Elasticsearch5.X

    最近做个项目需要使用到 Elasticsearch5 刚接触liunx 遇到了很多问题记录下 以这篇文章为基础 http://www.cnblogs.com/ShawnYuki/p/6818677.h ...

  5. CSIC_716_20191116【常用模块的用法 time ,datetime, random, os, sys, hashlib】

    import time import datetime import os import sys import random import hashlib time模块 时间戳(Timestamp) ...

  6. PHP缓存技术简单介绍

    一.数据缓存 这里所说的数据缓存是指数据库查询缓存,每次访问页面的时候,都会先检测相应的缓存数据是否存在,如果不存在,就连接数据库,得到数据,并把查询结果序列化后保存到文件中,以后同样的查询结果就直接 ...

  7. goconvey测试模块

    一.介绍 是一款针对Golang的测试框架,可以管理和运行测试用例,同时提供了丰富的断言函数,并支持很多 Web 界面特性. GoConvey 网站 : http://smartystreets.gi ...

  8. js--判断当前环境是否为iphonex环境

    /** * 判断是否是iphonex */ function getIsIphonex () { var u = navigator.userAgent; var isIOS = !!u.match( ...

  9. Http学习(一)

    HTTP 超文本传输协议 综述: HTTP(HyperText Transfer Protocol)是一套计算机通过网络进行通信的规则.计算机专家设计出HTTP,使HTTP客户(如Web浏览器)能够从 ...

  10. [kuangbin带你飞]专题一 简单搜索 - B - Dungeon Master

    #include<iostream> #include<cstdio> #include<string> #include<vector> #inclu ...