1048 Find Coins

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 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 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 (≤, 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 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

题意：

　　给出一个数组在这个数组中找出两个数，使其相加为m。

思路：

　　将数组排序，从数组的两头开始向中间收缩，直到两数相加等于m。

Code：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    vector<int> coins(n);
    for (int i = 0; i < n; ++i) cin >> coins[i];
    sort(coins.begin(), coins.end());
    int i = 0, j = coins.size() - 1;
    while (i < j) {
        if (coins[i] + coins[j] > m)
            j--;
        else if (coins[i] + coins[j] < m)
            i++;
        else
            break;
    }
    if (i >= j)
        cout << "No Solution" << endl;
    else
        cout << coins[i] << " " << coins[j] << endl;
    return 0;
}