UVA1152- 枚举 /二分查找】的更多相关文章

1514: Packs Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 61  Solved: 4[Submit][Status][Web Board] Description Give you n packs, each of it has a value v and a weight w. Now you should find some packs, and the total of these value is max, total of…
题目链接:https://vjudge.net/problem/POJ-3977 题意:给一个大小<=35的集合,找一个非空子集合,使得子集合元素和的绝对值最小,如果有多个这样的集合,找元素个数最少的. 思路:显然,可以用折半搜索,分别枚举一半,最大是2的18次方,复杂度能够满足.因为集合非空,枚举时考虑只在前一半选和只在后一半选的情况.对于前一半后一半都选的情况,把前一半的结果存下来,排序,枚举后一半的时候在前一半里二分查找最合适的即可. 思路不难,实现有很多细节,最开始用dfs写得一直wa,…
题目描述 Given a list of N integers with absolute values no larger than 10 15, find a non empty subset of these numbers which minimizes the absolute value of the sum of its elements. In case there are multiple subsets, choose the one with fewer elements.…
题目链接:http://poj.org/problem?id=3977 给你n个数,找到一个子集,使得这个子集的和的绝对值是最小的,如果有多种情况,输出子集个数最少的: n<=35,|a[i]|<=10e15 子集个数共有2^n个,所以不能全部枚举,但是可以分为两部分枚举: 枚举一半的所有情况,然后后一半二分即可: #include<iostream> #include<algorithm> #include<string.h> #include<st…
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 .…
Eqs Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 13955   Accepted: 6851 Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It i…
https://vjudge.net/problem/UVA-1152 题意:给定4个n元素集合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.问有多少种取法. 思路:直接暴力枚举的话是会超时的.可以选把a+b的值枚举出来存储,c和d的值也一样并排序,这样就可以在c和d中进行二分查找了. #include<iostream> #include<algorithm> using namespace std; + ; int n; int a[maxn]…
题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=679 解题报告:给定一个正整数的序列,和一个S,求长度最短的子序列,使它们的和大于或等于S.序列长度n <= 100000 很明显,如果枚举起点和终点的话,时间复杂度是O(n^3),不行.怎么能在O(1)时间求出一个子序列的和是多少呢,可以用另一个数组sum[i…
题意: 求正整数L和U之间有多少个整数x满足形如x=pk 这种形式,其中p为素数,k>1 分析: 首先筛出1e6内的素数,枚举每个素数求出1e12内所有满足条件的数,然后排序. 对于L和U,二分查找出小于U和L的最大数的下标,作差即可得到答案. #include <cstdio> #include <cmath> #include <algorithm> typedef long long LL; ; ; ]; ; LL a[maxn], cnt = ; void…
描述 http://poj.org/problem?id=3685 一个n*n的矩阵,(i,j)的值为i*i+100000*i+j*j-100000*j+i*j,求第m小的值. Matrix Time Limit: 6000MS   Memory Limit: 65536K Total Submissions: 5980   Accepted: 1700 Description Given a N × N matrix A, whose element in the i-th row and j…