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…

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,…

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 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…

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: 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: 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…

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…

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 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: 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…

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…

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…

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…

给出四个长度为n的数列a,b,c,d,求从这四个数列中每个选取一个元素后的和为0的方法数.n<=4000,abs(val)<=2^28. 考虑直接暴力,复杂度O(n^4).显然超时. # include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # in…

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 .…

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…

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…

题意 给你长度为n四个数列,每个数列选一个数使总和为4,有多少种选法(不同选法仅当起码有一个元素的下标不同) 输入 第一行,n 下面n行,每行四个数,代表ai,bi,ci,di 输出 选法数量 样例输入 6 -45 22 42 -16 -41 -27 56 30 -36 53 -37 77 -36 30 -75 -46 26 -38 -10 62 -32 -54 -6 45 样例输出 5 分析 1.傻逼算法:四重枚举 O(n4) 2.优化一点的傻逼算法:三重枚举+一重二分O(n3logn) 3.…

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…

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…

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…

  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…

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&…

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…