Find Coins (25)

思路：

500個面值，哈希存儲即可。注意從1開始算的話，注意M-i>500的情況。

代碼：

#include "iostream"
#include "vector"
#include "string"
#include "cstring"
#include "queue"
#include "map"
#include "cstdio"
#include <limits.h>
using namespace std;

int main()
{
	int N, M;
	cin >> N >> M;
	int lim = M / 2;
	vector<int> result(501, 0);
	for (int i = 0; i < N; i++)
	{
		int k;
		cin >> k;
		result[k]++;
	}
	int i = 1;
	for (; i <= M/2; i++)
	{
		if (M - i >= 500)
			continue;
		if (M % 2 == 0 && i == M / 2)
		{
			if (result[i] >= 2)
				break;
			else
				continue;
		}
		if (result[i] != 0 && result[M - i] != 0)
			break;
	}
	if (i == M/2 + 1)
		cout << "No Solution";
	else
		cout << i << ' ' << M - i;
}