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

idea

• 测试点1：m比i小的情况处理
• 小细节：i == m/2时，注意i的出现次数必须大于一次

solution

#include <stdio.h>
int main(){
	int n, m, d, hash[501] = {0};
	scanf("%d%d", &n, &m);
	for(int i = 0; i < n; i++){
		scanf("%d", &d);
		hash[d]++;
	}
	for(int i = 1; i < 501; i++){
		if(hash[i] && m > i && m - i <= 500){
			hash[i]--;
			if(hash[m-i]) {
				printf("%d %d", i, m - i);
				return 0;
			}
		}
	}
	printf("No Solution");
	return 0;
}