标签：1068 PAT Level int money coins index new Integer

1068 Find More Coins (30 分)

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 104 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 (≤10⁴, the total number of coins) and M (≤10², 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​≤V₂​≤⋯≤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

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
     public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

        String s = reader.readLine();
        String[] split = s.split(" ");
        int N = Integer.parseInt(split[0]); // 硬币总数
        int M = Integer.parseInt(split[1]); //待付金额

        String s1 = reader.readLine();
        String[] s2 = s1.split(" ");
        Integer[] coins = new Integer[N];          //拥有的硬币面额
        for (int i = 0; i < N; i++) {
            coins[i] = Integer.parseInt(s2[i]);
        }
        FindMoreCoins(M, N, coins);
    }

    public static void FindMoreCoins(int money, int total, Integer[] coins){
        int n = coins.length;
        int[] dp = new int[money + 1];
        boolean[][] flag = new boolean[total+1][money + 1];
        Arrays.fill(dp, -1);
        dp[0] = 0;
        Arrays.sort(coins, new Comparator<Integer>() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return o2 - o1;
            }
        });

        for (int i = 0; i < n; i++){
            int coin = coins[i];
            for (int j = money; j >= coin; j--){
                if (dp[j] <= dp[j - coin] + coin){
                    dp[j] =  dp[j - coin] + coin;
                    flag[i+1][j] = true;
                }
            }
        }
        if (dp[money] != money){
            System.out.println("No Solution");
        }else {
            ArrayList<Integer> arr = new ArrayList<>();
            int v = money, index = n;
            while(v > 0) {
                if(flag[index][v] == true) {
                    arr.add(coins[index-1]);
                    v -= coins[index-1];
                }
                index--;
            }
            for(int i = 0; i < arr.size(); i++) {
                if(i != 0) System.out.print(" ");
                System.out.print(arr.get(i));
            }
        }
    }
}

　　

 

   

标签：1068,PAT,Level,int,money,coins,index,new,Integer
来源： https://www.cnblogs.com/jasonXY/p/15171906.html

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