POJ 2366 Sacrament of the sum(哈希)

题目：2366

题意：从两个有序列表中寻找是否存在两个数加和等于10000

题解：哈希

AC代码：

#include<iostream>
using namespace std;

#define MAX 80000
#define mid 40000
#define target 10000
int a[MAX] = {0};

int main() {
	int n1, n2;
	int temp;
	cin >> n1;
	for (int i = 0; i < n1; i++) {
		cin >> temp;
		a[temp + mid] = 1;
	}
	cin >> n2;
	for (int i = 0; i < n2; i++)
	{
		cin >> temp;
		if (a[target - temp + mid]) {
			cout << "YES" << endl;
			return 0;
		}
	}
	cout << "NO" << endl;
	return 0;
}