题目

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: $N$ ( $\le 10^5$ , the total number of coins) and $M$ ( $\le 10^3$ , 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 $V_1 \le V_2$ . 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:

1
2

8 15
1 2 8 7 2 4 11 15

Sample Output 1:

1

4 11

Sample Input 2:

1
2

7 14
1 8 7 2 4 11 15

Sample Output 2:

1

No Solution

思路

可能是最简单的甲级题目之一了。

从诸多面值的硬币中找到正好凑齐交付金额的两枚硬币。鉴于面额的范围（1-500）很小，可以开一个数组记录不同面额硬币的数量，比较方便。

代码

Github最新代码，欢迎交流

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#include <stdio.h>

int main()
{
	int N, M, counts[501] = {0}, coin;

	scanf("%d %d", &N, &M);
	for (int i = 0; i < N; i++) {
		scanf("%d", &coin);
		counts[coin]++;
	}

	for (int i = 1; 2 * i - 1 < M; i++) {
		if ((i * 2 == M && counts[i] > 1)
		|| (i * 2 != M && M - i < 501 && counts[i] && counts[M - i])) {
			printf("%d %d", i, M - i);
			return 0;
		}
	}

	printf("No Solution");
	return 0;
}