Farey Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14291   Accepted: 5647 Description The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b)…

Farey Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17744   Accepted: 7109 Description The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b)…

[POJ 3581]Sequence 标签: 后缀数组 题目链接 题意 给你一串序列\(A_i\),保证对于$ \forall i \in [2,n],都有A_1 >A_i$. 现在需要把这个序列分成三段,并且将这三段分别翻转,求如何翻转使整个序列字典序最小.(每一段不能为空) 题解 首先可以确定第一段的位置. 注意到,\(A_1\)是最大的,所以我们就只用考虑怎样找到一个前缀使其翻转后的字典序最小. (假如不是的话,就可能找到两个前缀翻转之后,一个为另一个的前缀,无法解决) 这等价于翻转之后找…

Pro. 1 给定k个有序表,取其中前n小的数字.组成一个新表,求该表? 算法: 由于  a1[1] < a1[2] < a1[3] ... <a1[n] a2[1] < a2[2] < a2[3] ... < a2[n] ......... ak[1] < ak[2]<ak[3]...... < ak[n] 首先每个有序表的第一个元素入堆,然后最小元素出堆.该元素入新表L,相应线性表的下一个元素入堆. 例如:如果出堆得是a2[2],那么a2[3]入堆…

Sequence Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 6911   Accepted: 1543 Case Time Limit: 2000MS Description Given a sequence, {A1, A2, ..., An} which is guaranteed A1 > A2, ..., An,  you are to cut it into three sub-sequences and…

题目链接:http://poj.org/problem?id=2442 Time Limit: 6000MS Memory Limit: 65536K Description Given m sequences, each contains n non-negative integer. Now we may select one number from each sequence to form a sequence with m integers. It's clear that we ma…

Sequence Sum Possibilities Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5537   Accepted: 3641 Description Most positive integers may be written as a sum of a sequence of at least two consecutive positive integers. For instance, 6 = 1…

题目:http://poj.org/problem?id=2442 题意:给你n*m的矩阵,然后每行取一个元素,组成一个包含n个元素的序列,一共有n^m种序列, 让你求出序列和最小的前n个序列的序列和. 又是一个机智的题 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #in…

题目: http://poj.org/problem?id=2442 #include <stdio.h> #include <string.h> #include <queue> #include <algorithm> using namespace std; priority_queue<int>q; ][]; int main() { int t, n, m; scanf("%d", &t); while(t-…

[题目链接] http://poj.org/problem?id=3581 [题目大意] 给出一个数列,将这个数列分成三段,每段分别翻转,使得其字典序最小,输出翻转后的数列. [题解] 首先,第一个翻转点就是翻转后数列的最小后缀,注意由于一定要分成三段,则至少要剩下两个元素.难点主要是如何处理第二个翻转点,我们发现剩余的部分的每一种翻转拆分都是将两串翻转后剩余部分拼接在一起得到的串的子串,所以我们将剩余部分翻转,复制一份拼接在后面,求最小后缀即可. [代码] #include <cstdio>…