题意:问有没有数对(i,j)(0<=i<=j<n),使得a[i]-a[i+1]+...+(-1)^(j-i)a[j]为K. 解法:两种方法,枚举起点或者枚举终点. 先保存前缀和:a1-a2+a3....+/- an 枚举起点法: 设起点为x,实际是枚举x-1,分两种情况: 1.起点x为奇,那么就看有没有a[j]-a[x-1] = K的,即a[j] = a[x-1]+K.因为奇数位置的ai数符为正. 2.起点x为偶,那么就看有没有a[j]-(-K) = a[x-1],即a[j] = a[x…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5183 Negative and Positive (NP) Description When given an array $\left( {{a_0},{a_1},{a_2}, \cdots {a_{n - 1}}} \right)$ and an integer $K$, you are expected to judge whether there is a pair $(i,j)\ (0 \…

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−…

学到了以邻接表方式建立的hashmap 题意:给你一串数a和一个数k,都有正有负,问知否能找到一对数(i,j)(i<=j)保证a [i] - a [i+1] + a [i+2] - a [i+3]........(-1)^(j-i) a[j] 等于k 题解:想了很久才想出一个方法就是:记录前缀和,利用前缀和可以求所有可能性:对于每次求前缀和psum,psum[i]及其psum[i]-psum[比i小的](就是减去之前每次求出的前缀和)组成的小于n*n/2个数字就是总的可能出现的数(当然要处理一下…

题目链接: hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5183 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=570&pid=1002 题解: 前缀和+哈希 维护两张哈希表hash1,hash2,一张维护前缀和sum=a[0]-a[1]+a[2]-a[3]+...+(-1)^i*a[i],另一张维护-sum=-a[0]+a[1]-a[2]+a[…

根据奇偶开两个hash表来记录后缀和.注意set会被卡,要手写hash表. 具体见代码: #include <stdio.h> #include <algorithm> #include <string.h> using namespace std; + ; + ; typedef long long ll; struct hashmap { int head[HASH], nxt[N], size; ll state[N]; void init() { size =…

题意: 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<n) which makes that NP−sum(i,j) equals to K true. Here NP−sum(i,j)=ai−ai+1+ai+2+⋯+(−1)j−iaj 1≤n≤1000000,−1000000000≤ai≤1000000000…

Negative and Positive (NP) Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2177    Accepted Submission(s): 556 Problem Description When given an array (a0,a1,a2,⋯an−1) and an integer K, you are…

题目链接:HDOJ - 5183 题目分析 分两种情况,奇数位正偶数位负或者相反. 从1到n枚举,在Hash表中查询 Sum[i] - k ,然后将 Sum[i] 加入 Hash 表中. BestCoder比赛的时候我写了 STL map, 然后TLE... 注意: Hash负数的时候 % 了一个质数,得到的是负数还要 + Mod !! 代码 #include <iostream> #include <cstdio> #include <cstdlib> #includ…

传送门:Negative and Positive (NP) 题意:给定一个数组(a0,a1,a2,⋯an−1)和一个整数K, 请来判断一下是否存在二元组(i,j)(0≤i≤j<n)使得 NP−sum(i,j) 刚好为K.这里NP−sum(i,j)=ai−ai+1+ai+2+⋯+(−1)j−iaj. 分析:根据a[i]的i为奇偶进行hash,维护两种前缀和 1)i为奇数开头:sum=a[i]-a[i+1]+a[i+2]... 2)i为偶数开头:sum=a[i]-a[i+1]+a[i+2]...…