http://poj.org/problem?id=2785 题目大意: 给你四个数组a,b,c,d求满足a+b+c+d=0的个数 其中a,b,c,d可能高达2^28 思路: 嗯,没错,和上次的 HDU 1496 Equations hash(见http://blog.csdn.net/murmured/article/details/17596655)差不多,但是那题数据量小,hash值很好得到,不用取模运算. 而这题数据量很大.那就采用开散列的思想. hash值怎么选取? 上次我看的书中是建…

传送门 4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 20334   Accepted: 6100 Case Time Limit: 5000MS Description The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute…

4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 13069   Accepted: 3669 Case Time Limit: 5000MS Description The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how…

4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 25615   Accepted: 7697 Case Time Limit: 5000MS Description The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how…

4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 25675   Accepted: 7722 Case Time Limit: 5000MS Description The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how…

题目链接: https://cn.vjudge.net/problem/POJ-2785 The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we…

[题目链接] http://poj.org/problem?id=2785 [题目大意] 给出四个数组,从每个数组中选出一个数,使得四个数相加为0,求方案数 [题解] 将a+b存入哈希表,反查-c-d [代码] #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int N=4500,mod=1<<22; int n,a[N],b[N],c[…

Description The SUM problem can be formulated . In the following, we assume that all lists have the same size n . Input The first line of the input file contains the size of the lists n (). We then have n lines containing four integer values (with ab…

题目地址:http://poj.org/problem?id=2785 #include<cstdio> #include<iostream> #include<string.h> #include<algorithm> #include<math.h> #include<stdbool.h> #include<time.h> #include<stdlib.h> #include<set> #in…

思路: 如果用朴素的方法算O(n^4)超时,这里用折半二分.把数组分成两块,分别计算前后两个的和,然后枚举第一个再二分查找第二个中是否有满足和为0的数. 注意和有重复 #include<iostream> #include<algorithm> #include<cstring> #define ll long long using namespace std; const int N = 4000+5; int a[N],b[N],c[N],d[N]; int mp1…