L - Minimum Sum LCM Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 10791   题意:输入正整数n,<注意n=2^31-1是素数.结果是2^31已经超int.用long long,>找至少两个数,使得他们的LCM为n且要输出最小的和: 思路:既然LCM是n,那么一定是n的质因子组成的数,又要使和最小,那么就是ans+…

UVA.10791 Minimum Sum LCM (唯一分解定理) 题意分析 也是利用唯一分解定理,但是要注意,分解的时候要循环(sqrt(num+1))次,并要对最后的num结果进行判断. 代码总览 #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #define nmax 505 #define ll long long using namespace…

Minimum Sum Problem's Link ---------------------------------------------------------------------------- Mean: 给定n个整数,从中选出m个整数出来,使得这m个整数两两求(差的绝对值),并保证(差的绝对值)之和最小. analyse: 首先,要使得m个数(差的绝对值)之和最小,易知这m个数应该是连续的,所以先排序. 然后就是滑窗法了. 滑的时候如何维护滑块的sum呢? 如果我们选出的数是:…

Problem 1603 - Minimum Sum Time Limit: 2000MS   Memory Limit: 65536KB    Total Submit: 563  Accepted: 156  Special Judge: No Description There are n numbers A[1] , A[2] .... A[n], you can select m numbers of it A[B[1]] , A[B[2]] ... A[B[m]]  ( 1 <= B…

You are given an array a consisting of n integers a1, ..., an. In one operation, you can choose 2 elements ai and aj in which ai is divisible by aj and transform ai to aj. A number x is said to be divisible by a number y if x can be divided by y and…

You are given an array a consisting of n integers a1, ..., an. In one operation, you can choose 2 elements ai and aj in which ai is divisible by aj and transform ai to aj. A number x is said to be divisible by a number y if x can be divided by y and…

Problem 1603 - Minimum Sum Time Limit: 2000MS   Memory Limit: 65536KB   Total Submit: 623  Accepted: 178  Special Judge: No Description There are n numbers A[1] , A[2] .... A[n], you can select m numbers of it A[B[1]] , A[B[2]] ... A[B[m]]  ( 1 <= B[…

Minimum Sum 被这个题坑了一下午,原来只需找一个最中间的数即可,我以为是平均数. 题意:找一个数使得这个数和区间内所有数的差的绝对值最小.输出最小值. 开始用线段树来了一发果断T了,然后各种优化依然T不断.于是只能用划分树做.不过,其实熟悉划分树也是模板水题. 打一个前缀和,然后这个数肯定是这个区间从大到小排在最中间的.所以查询一遍将小于中间这个数的和加起来,然后得到将小于这个数的个数.分为前半段和后半段分别加起来即可. 手残将一个地方写错了,然后无限RE....最后一行一行debug…

http://www.practice.geeksforgeeks.org/problem-page.php?pid=166 Minimum sum partition Given an array, the task is to divide it into two sets S1 and S2 such that the absolute difference between their sums is minimum. Input: The first line contains an i…

题意:给定一个数组,有Q次的询问,每次询问的格式为(l,r),表示求区间中一个数x,使得sum = sigma|x - xi|最小(i在[l,r]之间),输出最小的sum. 思路:本题一定是要O(nlogn)或更低复杂度的算法.首先很容易得出这个x的值一定是区间(l,r)的中位数的取值,排序之后,也就是假设区间(l,r)长度为len ,则中位数就是该区间的第(r - l) / 2 - 1小的元素,求一个区间的第K小元素的算法很自然地会想到划分树, 而且划分树的查询复杂度为:O(logn),正好可…