4 Values whose Sum is 0 题目链接:https://cn.vjudge.net/problem/UVA-1152 ——每天在线,欢迎留言谈论. 题目大意: 给定4个n(1<=n<=4000)元素的集合 A.B.C.D ,从4个集合中分别选取一个元素a, b,c,d.求满足 a+b+c+d=0的对数. 思路: 直接分别枚举 a,b,c,d ,坑定炸了.我们先枚举 a+b并储存,在B.C中枚举找出(-c-d)后进行比较即可. 亮点: 由于a+b,中会有值相等的不同组合,如果使…
题意:给出n,四个集合a,b,c,d每个集合分别有n个数,分别从a,b,c,d中选取一个数相加,问使得a+b+c+d=0的选法有多少种 看的紫书,先试着用hash写了一下, 是用hash[]记录下来a[i]+b[j]的值, 如果a[i]+b[j]>0,hash[a[i]+b[j]]=1 如果a[i]+b[j]<0,hash[-(a[i]+b[j])]=-1 再用一个hash0[]去储存c[i]+d[j] 这样只需要满足hash[i]==1||hash0[i]==-1或者hash[i]==-1,…
题目:点击打开题目链接 思路:暴力循环显然会超时,根据紫书提示,采取问题分解的方法,分成A+B与C+D,然后采取二分查找,复杂度降为O(n2logn) AC代码: #include <bits/stdc++.h> using namespace std; ; int main() { ios::sync_with_stdio(false); cin.tie(); int T, n, ans; cin >> T; int A[maxn], B[maxn], C[maxn], D[ma…
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: 21370   Accepted: 6428 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…
摘要:中途相遇.对比map,快排+二分查找,Hash效率. n是4000的级别,直接O(n^4)肯定超,所以中途相遇法,O(n^2)的时间枚举其中两个的和,O(n^2)的时间枚举其他两个的和的相反数,然后O(logN)的时间查询是否存在. 首先试了下map,果断TLE //TLE #include<cstdio> #include<algorithm> #include<map> using namespace std; ; ][maxn]; map<int,in…
The SUM problem can be formulated as follows: given four lists A;B;C;D of integer values, computehow many quadruplet (a; b; c; d) 2 AB C D are such that a+b+c+d = 0. In the following, weassume that all lists have the same size n.InputThe input begins…
2017-08-01 21:29:14 writer:pprp 参考:http://blog.csdn.net/piaocoder/article/details/45584763 算法分析:直接暴力复杂度过高,所以要用二分的方法,分成两半复杂度就会大大降低: 题目意思:给定4个n(1<=n<=4000)元素的集合 A.B.C.D ,从4个集合中分别选取一个元素a, b,c,d.求满足 a+b+c+d=0的个数 代码如下: //首先将前两列任意两项相加得到数组x,再将后两列任意两项相加取反得到…
K - 4 Values whose Sum is 0 Crawling in process... Crawling failed Time Limit:9000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1152 Appoint description: System Crawler (2015-03-12) Description   The SUM problem c…
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: 17875 Accepted: 5255 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 man…
4 Values whose Sum is 0 Time Limit: 15000MS Memory Limit: 228000K Total Submissions: 19322 Accepted: 5778 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 man…
Problem UVA1152-4 Values whose Sum is 0 Accept: 794  Submit: 10087Time Limit: 9000 mSec Problem Description 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×B×C×D are…
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…
4 Values whose Sum is 0 Time Limit: 15000MS Memory Limit: 228000K Total Submissions: 26974 Accepted: 8133 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 man…
4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 29243   Accepted: 8887 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: 23757   Accepted: 7192 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:http://poj.org/problem?id=2785 题意是给你4个数列.要从每个数列中各取一个数,使得四个数的sum为0,求出这样的组合的情况个数. 其中一个数列有多个相同的数字时,把他们看作不同的数字. 做法是把前两个数列和的值存在一个数组(A)中 , 后两个数列的和存在另一个数组(B)中 , 数组都为n^2 . 然后将B数组sort一下 , 将A数组遍历 , 二分查找一下B数组中与A数组元素和为0的个数 . 有个注意的点是万一A数组都是0 , 而B数组都为0的情况(还有其…
题意: 要从四个数组中各选一个数,使得这四个数之和为0,求合法的方案数. 分析: 首先枚举A+B所有可能的值,排序. 然后枚举所有-C-D的值在其中用二分法查找. #include <cstdio> #include <algorithm> using namespace std; + ; int A[maxn], B[maxn], C[maxn], D[maxn], sum[maxn*maxn], cnt; int main() { //freopen("in.txt&…
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 显然中间相遇. 自己写了个hash处理一下冲突就可以了. [代码] /* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself? 3.Can you promise that the solution is right? At least,the main ideal 4.use the puts("&quo…
  Time Limit:15000MS     Memory Limit:228000KB     64bit IO Format:%I64d & %I64u Submit Status Description 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…
传送门:http://poj.org/problem?id=2785 Description 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,…
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 Description 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 assume that all lists have…
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 assume that all lists have the same size n .…
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…
题目大意:在4个都有n个元素的集合中,每个集合选出一个元素,使得4个数和为0.问有几种方案. 题目分析:二分.任选两组求和,剩下两组求和,枚举第一组中每一个和sum,在第二组和中查找-sum的个数,累加起来便得答案. 代码如下: # include<iostream> # include<cstdio> # include<vector> # include<cstring> # include<algorithm> using namespac…
找四个数的和为0 题目大意:给定四个集合,要你每个集合选4个数字,组成和为0 这题是3977的简单版,只要和是0就可以了 #include <iostream> #include <algorithm> #include <functional> #define MAX 4001 using namespace std; typedef long long LL_INT; ][MAX], set_sum1[MAX*MAX]; LL_INT *Binary_Lower_B…
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…