Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8403   Accepted: 3264 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…

题目连接 http://poj.org/problem?id=3061 Subsequence 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 given. Write a program to find the minimal len…

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 given. Write a program to find the minimal length of the subsequence of consecutive elements o…

[题目链接] http://poj.org/problem?id=3061 [题目大意] 给出S和一个长度为n的数列,问最短大于等于S的子区间的长度. [题解] 利用双指针获取每一个恰好大于等于S的子区间,更新答案即可. [代码] #include <cstdio> int T,a[100005]; int main(){ scanf("%d",&T); while(T--){ int n,S,s,h,t,ans; scanf("%d%d",&a…

http://poj.org/problem?id=3061 题目大意: 给定长度为n的整列整数a[0],a[1],--a[n-1],以及整数S,求出总和不小于S的连续子序列的长度的最小值. 思路: 方法一: 首先求出各项的和sum[i],这样可以在O(1)的时间内算出区间上的总和,这样,枚举每一个起点i,然后二分搜索出结果大于sum[i]+tot的最小下标.(tot是题目中的S) 总的时间为O(nlogn) 方法二: 设以a[s]开始的总和最初大于S时的连续子序列为a[s]+a[s+1]+--…

地址 http://poj.org/problem?id=3061 解法1 使用双指针 由于序列是连续正数 使用l r 表示选择的子序列的起始 每当和小于要求的时候 我们向右侧扩展 增大序列和 每当和大于等于要求的时候 我们将子序列左边的数字剔除 看能是在减少长度情况下 还能保持子序列和满足要求 这样在指定起点下的满足要求的最短子序列和都会被记录 然后在比较出最短长度的子序列 如图 代码 #include <iostream> #include <vector> #include…

题目链接: 传送门 Subsequence Time Limit: 1000MS     Memory Limit: 65536K 题目描述 给定长度为n的数列整数以及整数S.求出总和不小于S的连续子序列的长度的最小值.如果解不存在,则输出0. 思路 O(nlogn)算法 #include<cstdio> #include<iostream> #include<algorithm> using namespace std; int main() { int T; sca…

Subsequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12333 Accepted: 5178 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 gi…

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…

Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9050   Accepted: 3604 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…

Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18145   Accepted: 7751 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…

题目链接 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 given. Write a program to find the minimal length of the subsequence of consecutive eleme…

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 given. Write a program to find the minimal length of the subsequence of consecutive elements o…

Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14698   Accepted: 6205 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…

转自博客:http://blog.chinaunix.net/uid-24922718-id-4848418.html 尺取法就是两个指针表示区间[l,r]的开始与结束 然后根据题目来将端点移动,是一种十分有效的做法.适合连续区间的问题 poj3061 给定长度为n的数列整数a0,a1,a2,a3 ..... an-1以及整数S.求出综合不小于S的连续子序列的长度的最小值.如果解不存在,则输出0. 这里我们拿第一组测试数据举例子,即 n=10, S = 15, a = {5,1,3,5,10,7…

题目大意:给出长度为n的一个序列,给出一个数字S,求长度最短的序列和大于等于S的连续子序列,输出该长度,如果没有答案输出0. 题目思路:看数据范围,这道题就是卡时间的.我们可以用sum[i]记录前i项和,然后用二分优化查找过程.这样时间复杂度为 n*logn.具体看代码吧. #include<cstdio> #include<stdio.h> #include<cstdlib> #include<cmath> #include<iostream>…

<题目链接> 题目大意: 给你一段长度为n的整数序列,并且给出一个整数S,问你这段序列中区间之和大于等于S的最短区间长度是多少. 解题分析:本题可以用二分答案做,先求出前缀和,然后枚举区间长度,然后再判断其是否合法即可,复杂度$O(nlog(n))$.同时,尺取法也是一个不错的选择,通过不断的移动区间的头.尾指针来寻求答案,复杂度为 $O(n)$. 尺取法: #include <cstdio> #include <cstring> #include <algori…

题意 : 找出给定序列长度最小的子序列,子序列的和要求满足大于或者等于 S,如果存在则输出最小长度.否则输出 0(序列的元素都是大于 0 小于10000) 分析 : 有关子序列和的问题,都可以考虑采用先构造前缀和的方式来进行接下来的操作 ( 任意子序列的和都能由某两个前缀和的差表示 ). 二分做法 ==> 我们枚举起点,对于每一个起点 St 二分查找看看 St 后面有没有前缀和是大于或者等于 [ St的前缀和 ] + S 的,如果有说明从当前起点开始有一个终点使得起终之和是大于或者等于 S 的,…

题目传送门 /* 题意:求连续子序列的和不小于s的长度的最小值 尺取法:对数组保存一组下标(起点,终点),使用两端点得到答案 1. 记录前i项的总和,求[i, p)长度的最小值,用二分找到sum[p] - s[i] >= s的p 2. 除了O (nlogn)的方法,还可以在O (n)实现,[i, j)的区间求和,移动两端点,更新最小值,真的像尺取虫在爬:) */ #include <cstdio> #include <algorithm> #include <cstri…

北大ACM - POJ试题分类 -- By EXP 2017-12-03 转载请注明出处: by EXP http://exp-blog.com/2018/06/28/pid-38/ 相关推荐文: 旧版POJ分类目录: http://exp-blog.com/2018/06/10/pid-136/ ACM绝版资源公开( 参考书.模板.讲义.指导): http://exp-blog.com/2018/07/11/pid-1777/ ACM国家集训队论文集(1999-2009): http://ex…

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…

Problem 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 given. Write a program to find the minimal length of the subsequence of consecutive elements of th…

Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10172   Accepted: 4160 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…

Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22040   Accepted: 9404 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…

  和最短序列 题目大意:找出一个序列中比至少和S相等的最短子序列(连续的) 本来这道题可以二分法来做复杂度O(NlogN),也可以用一个类似于游标卡尺的方法O(N)来做 先来讲游标卡尺法: 因为子序列是连续的,所以我们只用维护这个序列的开头和结尾就行了,保证这个序列的和一定要大于S,如果从头到尾的和都没S大那就直接输出0就好,ans初始化为n+1 #include <iostream> #include <functional> #include <algorithm>…

二分法+前缀和法律 满足子序列长度的条件(0,n)之间,sum[x+i]-sum[i]从i元素开始序列长度x和.前缀和可在O(n)的时间内统计 sum[i]的值.再用二分找出满足条件的最小的子序列长度. #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<queue> #include&…

题目链接: http://poj.org/problem?id=3061 题目大意:找到最短的序列长度,使得序列元素和大于S. 解题思路: 两种思路. 一种是二分+前缀和.复杂度O(nlogn).有点慢. 二分枚举序列长度,如果可行,向左找小的,否则向右找大的. 前缀和预处理之后,可以O(1)内求和. #include "cstdio" #include "cstring" ],n,s,a,T; bool check(int x) { int l,r; ;i+x-&…

POJ 1401 Factorial Time Limit:1500MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu   Description The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceivers form the areas called cells (this ter…

Check the difficulty of problems Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4748   Accepted: 2078 Description Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually exp…

SudoKu Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu POJ 2676 Description Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the c…