Given several segments of line (int the X axis) with coordinates [Li , Ri ]. You are to choose the minimal amount of them, such they would completely cover the segment [0, M]. Input The first line is the number of test cases, followed by a blank line…
原题poj 2299:http://poj.org/problem?id=2299 题意,给你一个数组,去统计它们的逆序数,由于题目中说道数组最长可达五十万,那么O(n^2)的排序算法就不要再想了,归并排序~~~ 这里列出归并的解法: #include"iostream" using namespace std; const int maxn=500000+10; int T[maxn]; int a[maxn]; long long sum; void merge_sort(int…
Description   There are N<tex2html_verbatim_mark> marbles, which are labeled 1, 2,..., N<tex2html_verbatim_mark> . The N<tex2html_verbatim_mark> marbles are put in a circular track in an arbitrary order. In the top part of the track ther…
A permutation on the integers from 1 to n is, simply put, a particular rearrangement of these integers. Your task is to generate a given permutation from the initial arrangement 1, 2, 3, . . . , n using only two simple operations. •  Operation 1: You…
原题:poj3061 题意:给你一个数s,再给出一个数组,要求你从中选出m个连续的数,m越小越好,且这m个数之和不小于s 这是一个二分查找优化题,那么区间是什么呢?当然是从1到数组长度了.比如数组长度为10,你先找5,去枚举每一个区间为5的连续的数,发现存在这样的数,那么就可以继续往左找,反之则往右找,直到左右区间重合,即为正确答案,(有可能是右区间小于左区间,所以每次都应该求区间中点值) #include"iostream" #include"set" #incl…
原题:UVA11078 题意:给你一个数组,设a[],求一个m=a[i]-a[j],m越大越好,而且i必须小于j 怎么求?排序?要求i小于j呢.枚举?只能说超时无上限.所以遍历一遍数组,设第一个被减数为a[0],之后遇到比a[0]大的数就更新它,再拿这个被减数去减数组中的每一个元素,同时也要不断地更新这个m值. #include"iostream" #include"set" #include"cstring" #include"cst…
Description Shaass has n books. He wants to make a bookshelf for all his books. He wants the bookshelf's dimensions to be as small as possible. The thickness of the i-th book is ti and its pages' width is equal to wi. The thickness of each book is ei…
UVA10382 :http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=21419 只能说这道题和D题是一模一样的,不过要进行转化,这道题有一道坑,方法一需要使用scanf()输入,还需判断输入的正确性,即scanf()==3..... 方法一: #include"iostream" #include"cstdio" #include"algorithm" #include&qu…
原题 UVA10020  :http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=19688 经典的贪心问题,区间上贪心当然是右区间越右越好了,当然做区间要小于等于当前待覆盖的位置,我们可以用一个变量begin代表当前待覆盖的区间的左值,每一次贪心成功就更新begin的值,然后再进行类似的重复动作 我之前写的版本.两次排序. #include"iostream" #include"cstdio" #i…
Description   John Doe is a famous DJ and, therefore, has the problem of optimizing the placement of songs on his tapes. For a given tape and for each song on that tape John knows the length of the song and the frequency of playing that song. His pro…