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…

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: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…

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&…

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…

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…

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: 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…

[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…

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&…

题目链接:https://vjudge.net/problem/UVA-1152 这题题意就是在四个集合内.每个集合分别里挑一个数a,b,c,d,求a+b+c+d=0有多少种选法. 暴力的话就是四重循环,复杂度O(n^4)容易超时.稍微优化一点就是分别枚举其中三个集合里的元素a+b+c,然后用二分查找剩下那个集合里存不存在a+b+c的相反数,复杂度O(n^3*logn)勉强还行. 不过可以枚举a+b,放入map中,然后枚举-c-d,看看-c-d有多少种方法能写成a+b.复杂度O(n^2*logn…

  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…

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…

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…