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…

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,…

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 such that a + b + c + d = . In the following, we assume that all lists have the same size n. Inp…

题意:从四个集合各选一个数,使和等于0,问有多少种选法. 分析:求出来所有ai + bi,在里面找所有等于ci + di的个数. #pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include&…

问题分析 首先枚举a和b, 把所有a+b记录下来放在一个有序数组,然后枚举c和d, 在有序数组中查一查-c-d共有多少个.注意这里不可以直接用二分算法的那个模板,因为那个模板只能查找是否有某个数,一旦找到便退出.利用lower_bound,upper_bound比较方便,这两个函数就是用二分实现的,二者之差就是相等的那部分. 代码 #include<bits/stdc++.h> using namespace std; typedef long long LL; LL T, n; const…

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…

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…

题目链接: https://cn.vjudge.net/problem/UVA-1152 题目大意: 给出4个数组,每个数组有n个数,问有多少种方案在每个数组中选一个数,使得四个数相加为0. n <= 4000 解题思路: 两重循环求出a + b的所有情况 两重循环求出-c - d的所有情况 枚举a+b的值,在-c-d里面找相同值的数目即可. #include<cstdio> #include<cstring> #include<iostream> #includ…

题意:给定4个N元素几个A,B,C,D,要求分别从中选取一个元素a,b,c,d使得a+b+c+d=0.问有多少种选法.(N≤4000,D≤2^28) 解法:首先我们从最直接最暴力的方法开始思考:四重循环O(n^4)枚举:三重循环枚举,把剩下的一个集合排序后二分查找,O(n^3 log n).在进一步想,运用"中途相遇法":从两个不同的方向来解决问题,最后"汇集"到一起的方法.(有类似于"双向广度优先搜索"的思想)通过两重循环枚举出A,B两个集合中…

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: 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…

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: 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…

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…

G - 4 Values whose Sum is 0 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…

4 Values whose Sum is 0 Time Limit: 15000MS   Memory Limit: 228000K Total Submissions: 18221   Accepted: 5363 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…

中途相遇法,这题目总结后我感觉和第一篇博客很像,他们都取了中间,也许这就是二分的魅力吧 这题题意就是从ABCD四个集合中选四个元素使他们的和为0 题意很简单,但是实现起来很容易超时,不能一个一个枚举 然后就想到了枚举两个,那么新生成EF两个集合大小为n方,然后如何查找? 最容易想到的就是用map标记,但汝佳说了这会超时,我不信,试了一发,超时了!!! 除了用map还能用什么呢,联想到之前那个UVA对称的题目,我们可以sort一下,然后查找e的相反数时候,二分查找-e,记住,找到还不行,有可能存在…

用中途相遇法的思想来解题.分别枚举两边,和直接暴力枚举四个数组比可以降低时间复杂度. 这里用到一个很实用的技巧: 求长度为n的有序数组a中的数k的个数num? num=upper_bound(a,a+n,k)-lower_bound(a,a+n,k); #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include&l…

这道题要逆向思维, 就是求出答案的一部分, 然后反过去去寻找答案存不存在. 其实很多其他题都用了这道题目的方法, 自己以前都没有发现, 这道题专门考这个方法.这个方法可以没有一直往下求, 可以省去很多时间.紫书里面把这叫做中途相遇法,双向广搜有点这个方法的味道.这里用到了二分查找, 总的时间复杂度是n的二次方乘logn #include<cstdio> #include<vector> #include<algorithm> #define REP(i, a, b) f…

uva 6757 Cup of CowardsCup of Cowards (CoC) is a role playing game that has 5 different characters (Mage, Tank, Fighter,Assassin and Marksman). A team consists of 5 players (one from each kind) and the goal is to kill amonster with L life points. The…

暴力n的四次方, 然而可以用中途相遇法的思想, 分左边两个数和右边两个数来判断, 最后合起来判断. 一边是n平方logn, 合起来是n平方logn(枚举n平方, 二分logn) (1)两种比较方式是相反的, 所以第二次可以直接把数组倒过来做, 代码可以省很多. (2) 我们现在来讨论3 1 4 2这种情况(1最小, 2次小以此类推) 大家观察可以发现, 中间两个数字刚好是最大和最小.所以我们可以枚举中间两个数, 往两边找. 先看1, 我们可以预处理出每一个数左侧比它大的数字有哪些.然后找到1的时…

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…