1048 Find Coins java

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 could only use exactly two coins to pay the exact amount. Since she has as many as 10​5​​ 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 two 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 (≤10​5​​, the total number of coins) and M (≤10​3​​, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the two face values V​1​​ and V​2​​ (separated by a space) such that V​1​​+V​2​​=M and V​1​​≤V​2​​. If such a solution is not unique, output the one with the smallest V​1​​. If there is no solution, output No Solution instead.

Sample Input 1:

8 15 1 2 8 7 2 4 11 15

Sample Output 1:

4 11

Sample Input 2:

7 14 1 8 7 2 4 11 15

Sample Output 2:

No Solution import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class Main { /** * 题意就是输入 n, m * 然后输入n个数，问在n个数中有没有2个数的和为m * 这个题目可以用二分先sort一下，然后二分查找， * 下面的代码用的是散列法 * 有2组数据超时 * * */ public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String inputString = br.readLine(); int m = Integer.parseInt(inputString.split("\\s+")[1]); String[] eachNumString = br.readLine().split("\\s+"); int[] counts = new int[1000+11]; for (int i=0; i<eachNumString.length; ++i) { int num = Integer.parseInt(eachNumString[i]); counts[num]++; } boolean tag = true; for (int i=1; i<m; ++i) { if (counts[i]>0&&counts[m-i]>0) { if (i==m-i&&counts[i]<=1) { continue; } System.out.println(i+" "+(m-i)); tag = false; break; } } if (tag) { System.out.println("No Solution"); } } }