HDU 5183 Negative and Positive (NP) (手写哈希)

题目链接：HDU 5183

Problem Description

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

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 \((a_0,a_1,a_2,?a_{n?1})\) separated by exact one space.

[Technical Specification]

All input items are integers.

\(0<T≤25,1≤n≤1000000,?1000000000≤a_i≤1000000000,?1000000000≤K≤1000000000\)

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.

Hint

If input is huge, fast IO method is recommended.

Source

BestCoder Round #32

Solution

题意

给定长度为 \(n\) 的数组，问是否存在一段区间其加减交错的和等于 \(k\)。

思路

对该数组求前缀和然后哈希就行。

不过这题不同的 set 过不了。

所以要手写哈希。

当然 unordered_set 也可以过。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 7;

struct HashMap {
    int head[maxn], next[maxn];
    ll num[maxn];
    int tot;
    inline void init() {
        tot = 0;
        memset(head, -1, sizeof(head));
    }
    inline void insert(ll val) {
        int h = abs(val) % maxn;
        num[tot] = val, next[tot] = head[h], head[h] = tot++;
    }
    inline bool find(ll val) {
        int h = abs(val) % maxn;
        for(int i = head[h]; ~i; i = next[i]) {
            if(num[i] == val) {
                return true;
            }
        }
        return false;
    }
} hashmap;

ll arr[maxn], sum[maxn];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    int kase = 0;
    while(T--) {
        hashmap.init();
        ll n, k;
        cin >> n >> k;
        for(int i = 1; i <= n; ++i) {
            cin >> arr[i];
            sum[i] = sum[i - 1] + (i & 1? arr[i]: -arr[i]);
        }
        int flag = 0;
        for(int i = n; i; --i) {
            hashmap.insert(sum[i]);
            if(i & 1) {
                if(hashmap.find(sum[i - 1] + k)) {
                    flag = 1;
                    break;
                }
            } else {
                if(hashmap.find(sum[i - 1] - k)) {
                    flag = 1;
                    break;
                }
            }
        }
        cout << "Case #" << ++kase << ": " << (flag? "Yes." : "No.") << endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/wulitaotao/p/11823679.html