hdu5183---Negative and Positive (NP)(HASHMAP)

最新推荐文章于 2021-06-06 22:45:40 发布

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本文提供了一道题目HDU 5183的解题思路及C++实现代码。该题目要求判断是否存在一对下标使得数组中元素的累加差值为给定整数K。通过使用哈希表结构来存储中间结果，实现了高效的查找和插入操作。

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Problem Description
When given an array (a0,a1,a2,⋯an−1) and an integer K, you are expected to judge whether there is a pair (i,j)(0≤i≤j

/*************************************************************************
    > File Name: hdu5183.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年03月08日 星期日 19时55分52秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int maxn = 1000007;

int A[1000010];

struct HASHMAP
{
    int head[maxn], next[maxn], size;
    LL sta[maxn];

    void init ()
    {
        size = 0;
        memset (head, -1, sizeof(head));
    }

    bool check (LL val)
    {
        int h = (val % maxn + maxn) % maxn;
        for (int i = head[h]; ~i; i = next[i])
        {
            if (val == sta[i])
            {
                return true;
            }
        }
        return false;
    }

    int insert (LL val)
    {
        int h = (val % maxn + maxn) % maxn;
        for (int i = head[h]; ~i; i = next[i])
        {
            if (val == sta[i])
            {
                return true;
            }
        }
        sta[size] = val;
        next[size] = head[h];
        head[h] = size++;
        return false;
    }
}H1, H2;

template <class T>
bool scan_d (T &ret)
{
    char c;
    int sgn;
    T bit = 0.1;
    if (c = getchar(), c == EOF)
    {
        return 0;
    }
    while (c != '-' && c != '.' && (c < '0' || c > '9'))
    {
        c = getchar ();
    }
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar (), c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0');
    }
    if (c == ' ' || c == '\n')
    {
        ret *= sgn;
        return 1;
    }
    while (c = getchar(), c >= '0' && c <= '9')
    {
        ret += (c - '0') * bit, bit /= 10;
    }
    ret *= sgn;
    return 1;
}

int main ()
{
    int t;
    int icase = 1;
    scanf("%d", &t);
    while (t--)
    {
        H1.init();
        H2.init();
        int n;
        scanf("%d", &n);
        LL k;
        scanf("%lld", &k);
        H1.insert (0);
        LL s = 0;
        bool flag = 0;
        for (int i = 0; i < n; ++i)
        {
            scan_d(A[i]);
        }
        for (int i = 0; i < n; ++i)
        {
            if (i & 1)
            {
                s -= A[i];
            }
            else
            {
                s += A[i];
            }
            if (i & 1)
            {
                H1.insert (s);
            }
            else
            {
                H2.insert (s);
            }
            if (H1.check (s - k) || H2.check (s + k))
            {
                flag = 1;
                break;
            }
        }
        printf("Case #%d: ", icase++);
        if (flag)
        {
            printf("Yes.\n");
        }
        else
        {
            printf("No.\n");
        }
    }
    return 0;
}