Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13955   Accepted: 5896 Description A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) ar…

题目传送门 /* 尺取法:先求出不同知识点的总个数tot,然后以获得知识点的个数作为界限, 更新最小值 */ #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MA…

Jessica's Reading Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6001   Accepted: 1800 Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent littl…

Jessica's Reading Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7467   Accepted: 2369 Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent littl…

http://poj.org/problem?id=3320 题意:给出一串数字,要求包含所有数字的最短长度. 思路: 哈希一直不是很会用,这道题也是参考了别人的代码,想了很久. #include<iostream> #include<algorithm> #include<string> #include<cstring> using namespace std; ; int n; int len; //开散列法,也就是用链表来存储,所以下面的len是从P…

Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The au…

Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The author of that…

Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The au…

地址 http://poj.org/problem?id=3320 解答 使用双指针 在指针范围内是否达到要求 若不足要求则从右进行拓展  若满足要求则从左缩减区域 代码如下  正确性调整了几次 然后被输入卡TLE卡了很久都没意识到......... #include <iostream> #include <map> #include <set> #include <algorithm> #include <assert.h> #include…

Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The au…

题意:给定一个序列,求一个最短区间,使得这个区间包含所有的种类数. 析:最近刚做了几个滑动窗口的题,这个很明显也是,肯定不能暴力啊,时间承受不了啊,所以 我们使用滑动窗口来解决,要算出所有的种数,我用set来计算的,当然也可以用别的, 由于要记录种类数,所以使用map来记录,删除和查找方便,说到这,这不就是水题了么. 代码如下: #include <iostream> #include <cstdio> #include <algorithm> #include <…

题目: 解法:定义左索引和右索引 1.先让右索引往右移,直到得到所有知识点为止: 2.然后让左索引向右移,直到刚刚能够得到所有知识点: 3.用右索引减去左索引更新答案,因为这是满足要求的子串. 4.不断重复1,2,3.直到搜索到最后,不论怎样都获得不了所有的知识点时跳出. 代码: #include <iostream> #include <algorithm> #include <stdio.h> #include <vector> #include <…

Jessica's Reading Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12944   Accepted: 4430 Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent litt…

jessica's Reading PJroblem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9134   Accepted: 2951 Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent litt…

A - Jessica's Reading Problem POJ - 3320 Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a…

Jessica's Reading Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17562   Accepted: 6099 Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent litt…

Jessica's Reading Problem 题目大意:Jessica期末考试临时抱佛脚想读一本书把知识点掌握,但是知识点很多,而且很多都是重复的,她想读最少的连续的页数把知识点全部掌握(知识点都在书上,每一页都是一个知识点) 这一题可以用3061的游标卡尺法,我们可以先数数书上倒到底有多少个知识点,因为知识点都是序数,我们可以用二分法直接找到O(PlogP),这里可以采用set模板直接偷懒了,然后我们就可以用游标卡尺法了,因为所有知识点都要出现一次,所以我们统计新的知识点的出现就好了,最…

Jessica's Reading Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12346   Accepted: 4199 Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent litt…

[POJ3320]Jessica's Reading Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13095   Accepted: 4495 Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has s…

POJ3320 Jessica's Reading Problem set用来统计所有不重复的知识点的数,map用来维护区间[s,t]上每个知识点出现的次数,此题很好的体现了map的灵活应用 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #include <vector> #include &…

Jessica's Reading Problem——POJ3320 题目大意: Jessica 将面临考试,她只能临时抱佛脚的在短时间内将课本内的所有知识点过一轮,课本里面的P个知识点顺序混乱,而且重复. 要求找出课本的连续页数,这些页数满足:包含全部知识点,且是包含全部知识点的页数区间里面最短的. 尺取法解题. import java.util.HashMap; import java.util.HashSet; import java.util.Iterator; import java.…

Jessica's Reading Problem Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 15224 Accepted: 5241 Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little t…

https://vjudge.net/problem/POJ-3320 尺取法,要想好组织方式. 又被卡了cin.. #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map> #include<…

题意:n页书,然后n个数表示各个知识点ai,然后,输出最小覆盖的页数. #include<iostream> #include<cstdio> #include<set> #include<map> using namespace std; ; int num[maxn]; int main(){ int n; set<int>ss; map<int, int>mm; scanf("%d", &n); ;…

Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The au…

Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The au…

这两道题都是用的尺取法.尺取法是<挑战程序设计竞赛>里讲的一种常用技巧. 就是O(n)的扫一遍数组,扫完了答案也就出来了,这过程中要求问题具有这样的性质:头指针向前走(s++)以后,尾指针(t)要么不动要么也往前走.满足这种特点的就可以考虑尺取法. poj3061 比较简单,也可以用二分做,时间复杂度O(n*logn).用尺取法可以O(n)解决. #include<iostream> #include<cstdio> #include<cstdlib> #i…

Bryce1010模板 #include <stdio.h> #include <string.h> #include <stdlib.h> #include <algorithm> #include <iostream> #include <queue> #include <stack> #include <vector> #include <map> #include <set> #…

题目链接: 传送门 Sum of Consecutive Prime Numbers Time Limit: 1000MS     Memory Limit: 65536K Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive intege…

Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9236   Accepted: 3701 Description A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are…