Jessica's Reading Problem 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 ve…

1.POJ 3320 2.链接:http://poj.org/problem?id=3320 3.总结:尺取法,Hash,map标记 看书复习,p页书,一页有一个知识点,连续看求最少多少页看完所有知识点 必须说,STL够屌.. #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio>…

题目链接: http://poj.org/problem?id=3320 题目大意:一本书有P页,每页有个知识点,知识点可以重复.问至少连续读几页,使得覆盖全部知识点. 解题思路: 知识点是有重复的,因此需要统计不重复元素个数,而且需要记录重复个数. 最好能及时O(1)反馈不重复的个数.那么毫无疑问,得使用Hash. 推荐使用map,既能Hash,也能记录对于每个key的个数. 尺取的思路: ①不停扩展R,并把扫过知识点丢到map里,直到map的size符合要求. ②更新结果. ②L++,map…

传送门:Problem 3320 参考资料: [1]:挑战程序设计竞赛 题意: 一本书有 P 页,每页都有个知识点a[i],知识点可能重复,求包含所有知识点的最少的页数. 题解: 相关说明: 设以a[start]开始的最初包含所有知识点的最少连续子序列为a[start,....,end]; mymap[ a[i] ] : 知识点 a[i] 在当前最少连续子序列中出现的次数. (1):求出所需复习的知识点总个数. (2):求出最先包含所有知识点的最少页数a[start,........,end].…

Description Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration (that must be going through a defiant phase: "But I want to use feet, not meters!"). Each signal seems to…

Boring count Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 932    Accepted Submission(s): 382 Problem Description You are given a string S consisting of lowercase letters, and your task is cou…

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…

poj3061 给定一个序列找出最短的子序列长度,使得其和大于等于S 那么只要用两个下标,区间和小于S时右端点向右移动,区间和大于S时左端点向右移动,在这个过程中更新Min #include <cstdio> #include <algorithm> #include <cstring> #define MAX 100005 #define LL long long #define INF 0x3f3f3f3f using namespace std; LL a[];…

[题目大意] 给出一个整数列,求一段子序列之和最接近所给出的t.输出该段子序列之和及左右端点. [思路] ……前缀和比较神奇的想法.一般来说,我们必须要保证数列单调性,才能使用尺取法. 预处理出前i个数的前缀和,和编号i一起放入pair中,然而根据前缀和大小进行排序.由于abs(sum[i]-sum[j])=abs(sum[j]-sum[i]),可以忽视数列前缀和的前后关系.此时,sum[r]-sum[l]有单调性. 因此我们可以先比较当前sum[r]-sum[l]与t的差,并更新答案. 如果当…

Graveyard Design Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 6107   Accepted: 1444 Case Time Limit: 2000MS Description King George has recently decided that he would like to have a new design for the royal graveyard. The graveyard m…