hdu 5183 Negative and Positive (NP)

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5183  

Negative and Positive (NP)

Description

When given an array $left( {{a_0},{a_1},{a_2}, cdots {a_{n - 1}}} ight)$ and an integer $K$, you are expected to judge whether there is a pair $(i,j) (0 leq i leq j < n)$ which makes that $NP−sum(i,j)$ equals to $K$ true. Here $NP-sum(i,j)={a_i}{ m{ - }}{a_{i{ m{ + 1}}}}{ m{ + }}{a_{i{ m{ + }}2}}{ m{ + }} cdots { m{ + ( - 1}}{{ m{)}}^{j - i}}{a_j}$

Input

Multi test cases. In the first line of the input file there is an integer $T$ indicates the number of test cases.
In the next $2∗T$ lines, it will list the data for each test case.
Each case occupies two lines, the first line contain two integers $n$ and $K$ which are mentioned above.
The second line contain $left( {{a_0},{a_1},{a_2}, cdots {a_{n - 1}}} ight)$separated by exact one space.
[Technical Specification]
All input items are integers.
$0 < T leq 25,1 leq n leq 1000000,-1000000000 leq ai leq 1000000000,-1000000000 leq K leq 1000000000$

Output

For each case，the output should occupies exactly one line. The output format is Case #id: ans, here id is the data number starting from 1; ans is “Yes.” or “No.” (without quote) according to whether you can find $(i,j)$ which makes $PN−sum(i,j)$ equals to $K$.
See the sample for more details.

Sample Input

2
1 1
1
2 1
-1 0

Sample Output

Case #1: Yes.
Case #2: No.

哈希大法好。。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<set>
using std::set;
using std::sort;
using std::pair;
using std::swap;
using std::queue;
using std::multiset;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 1000007;
const int INF = 0x3f3f3f3f;
typedef long long ll;
struct Hash_Set {
	ll num[N << 1];
	int tot, head[N], next[N];
	inline void init() {
		tot = 0, cls(head, -1);
	}
	inline void insert(ll val) {
		int u = abs(val) % N;
		num[tot] = val, next[tot] = head[u], head[u] = tot++;
	}
	inline bool find(ll val) {
		int u = abs(val) % N;
		for (int i = head[u]; ~i; i = next[i]) {
			if (num[i] == val) return true;
		}
		return false;
	}
}hash;
ll arr[N], sum[N];
int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w+", stdout);
#endif
	ll k;
	int t, n, c = 1;
	scanf("%d", &t);
	while (t--) {
		hash.init();
		scanf("%d %lld", &n, &k);
		for (int i = 1; i <= n; i++) scanf("%lld", &arr[i]);
		for (int i = 1; i <= n; i++) {
			sum[i] = sum[i - 1] + (i & 1 ? arr[i] : -arr[i]);
		}
		bool f = false;
		for (int i = n; i > 0; i--) {
			hash.insert(sum[i]);
			if (f) break;
			if (i & 1) {
				if (hash.find(sum[i - 1] + k)) {
					f = true;
					break;
				}
			} else {
				if (hash.find(sum[i - 1] - k)) {
					f = true;
					break;
				}
			}
		}
		printf("Case #%d: %s
", c++, f ? "Yes." : "No.");
	}
	return 0;
}