1.牛牛换零钱,求有多少种方式
输入:
第一行为询问数t.(也就是有t组数据)
第二行有两个数n和k,n为零钱的面值种类数,k为要换的零钱
第三行有n个数字,表示n个面值
例如:
1
3 5
1 2 5
输出
4
也就是有4种零钱组合的方式:11111,1112,122,5
就是说要牛牛要换1张钱,银行有3种面值的零钱可供选择,要换的钱的面值为5,三种面值分布为1、2和5
所以代码如下,仅供参考:
import java.util.ArrayList; import java.util.Collection; import java.util.HashMap; import java.util.Iterator; import java.util.Map; import java.util.Scanner; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub Scanner sc = new Scanner(System.in); while (sc.hasNextLine()) { int t = sc.nextInt(); Map<Integer, java.util.List<Integer>> map = new HashMap<Integer, java.util.List<Integer>>(); for (int i = 0; i < t; i++) { java.util.List<Integer> mylist = new ArrayList<>(); int n = sc.nextInt(); mylist.add(n); int k = sc.nextInt(); mylist.add(k); for (int s = 0; s < n; s++) { int temp = sc.nextInt(); mylist.add(temp); } map.put(i, mylist); } Collection<java.util.List<Integer>> value = map.values(); Iterator<java.util.List<Integer>> iter = value.iterator(); while (iter.hasNext()) { java.util.List<Integer> list = iter.next(); find(list); } } sc.close(); } private static void find(java.util.List<Integer> list) { // TODO Auto-generated method stub int len = list.size(); int[] num = new int[len]; for (int i = 0; i < list.size(); i++) { num[i] = list.get(i); } int n = num[0]; int k = num[1]; int[] values = new int[n]; for (int i = 0; i < values.length; i++) { values[i] = num[i + 2]; } solution(values, k); } private static void solution(int[] value, int k) { if (value == null || value.length == 0 || k < 0) { System.out.println(0); } int[] dp = new int[k + 1]; for (int j = 0; value[0] * j <= k; j++) { dp[value[0] * j] = 1; } for (int i = 1; i < value.length; i++) { for (int j = 1; j <= k; j++) { dp[j] += j - value[i] >= 0 ? dp[j - value[i]] : 0; } } System.out.println(dp[k] % 100000007); } }
1.牛牛卖青草
牛牛的邻居们围城了一个圈,相邻的两户人家,牛牛只能卖给一家,也就是当牛牛卖给第N户邻居时,不能再卖给第N-1户和第N+1户邻居。问牛牛最多能卖出去多少?
输入:
第一行为询问数t
第二行为牛牛的邻居数n
第三行为牛牛的邻居们需要的青草的数目
例如:
2
4
8 9 2 8
2
10 100
也就是有两组测试数据,第一组数据中有4个邻居,所需青草数为8,9,2,8;第二组数据有2个邻居,所需青草数为10,100
输出:
17
100
第一组邻居,牛牛卖了17,第二组邻居牛牛卖了100;
代码如下:
import java.util.ArrayList; import java.util.Collection; import java.util.HashMap; import java.util.Iterator; import java.util.LinkedList; import java.util.List; import java.util.Map; import java.util.Scanner; public class Main{ public static void main(String[] args) { // TODO Auto-generated method stub Scanner sc = new Scanner(System.in); while (sc.hasNext()) { int t = sc.nextInt(); Map<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>(); for (int i = 0; i < t; i++) { List<Integer> templist = new ArrayList<>(); int n = sc.nextInt(); templist.add(n); for (int j = 0; j < n; j++) { int temp = sc.nextInt(); templist.add(temp); } map.put(i, templist); } Collection<List<Integer>> values = map.values(); Iterator<List<Integer>> iter = values.iterator(); while (iter.hasNext()) { List<Integer> goallist = iter.next(); find(goallist); } } sc.close(); } private static void find(List<Integer> goallist) { // TODO Auto-generated method stub int len = goallist.size(); int[] num = new int[len]; for (int i = 0; i < goallist.size(); i++) { num[i] = goallist.get(i); } int n = num[0]; int[] value = new int[n]; for (int i = 0; i < value.length; i++) { value[i] = num[i+1]; } solution(n,value); } private static void solution(int n, int[] value) { // TODO Auto-generated method stub if (n==1||value.length==1) { System.out.println(value[0]); }else{ int max = 0; for (int i = 0; i < value.length; i++) { LinkedList<Integer> link = new LinkedList<>(); for (int j = i; j < value.length; j++) { link.add(value[j]); } for (int j = 0; j < i; j++) { link.add(value[j]); } int count = 0; int k = value.length/2; while (k>0) { int temp = link.poll(); count = count+temp; link.poll(); k = k - 1; } if (count>=max) { max = count; } } System.out.println(max); } } }(以上代码均通过了题目中的测试用例,但未在线检验,另外希望各位大神、大牛积极批评指正,优化代码!)