Subsequence POJ - 3061 给定长度为n的数列整数a0,a1,a2-an-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,如果解不存在输出0. 反复推进区间的开头和末尾,来求取满足条件的最小区间的方法称为取尺法. #include <cstdio> #include <iostream> #include <cmath> #include <algorithm> #include <string> #include &…

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…

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…

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: 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…

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…

https://vjudge.net/problem/POJ-3061 尺取发,s和t不断推进的算法.因为每一轮s都推进1所以复杂度为O(n) #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #define lson l, m,…

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…

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 the sequen…

尺取法就是反复推进区间的开头和结尾,来求满足条件的最下区间. poj3061 http://poj.org/problem?id=3061 给定一个都是正整数的序列,要我们求总和不小于S的连续子序列的长度的最小值 如果序列   是总和最迟大于S的连续子序列 那么  所以只有加上, 从开始的连续子序列才有可能大于S 所以从开始的总和最初大于S的连续子序列是则一定有 #include <stdio.h> #include <string.h> #include <stdlib.h…

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 the sequen…

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…

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…

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…

---恢复内容开始--- Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10487   Accepted: 4337 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 < 10…

题目大意:从给定序列里找出区间和大于等于S的最小区间的长度. 前阵子在zzuli OJ上见过类似的题,还好当时补题了.尺取法O(n) 的复杂度过掉的.尺取法:从头遍历,如果不满足条件,则将尺子尾 部增加,若满足条件,则逐渐减少尺子头部直到不满足条件为止,保存 尺子长度的最小值(尾部-头部+1)即可. 理论上累计区间和+二分查找的暴力也能过. 代码如下: #include <stdio.h> #include <algorithm> #include <string.h>…

<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…

POJ3061-Subsequence [题目大意] 给定长度微n的数列整数及整数s.求出总和不小于s的连续子序列的长度的最小值.如果节不存在,则输出0. [思路] 尺取法五分钟裸裸裸~刷水刷出了罪恶感:( 基本做法:设置l和r代表当前区间[l,r],若S(l,r)<s,则 r++.若S(l,r)≥s,则 l++,直至S(l,r)<s.如果当前S(l,r)<s且r=n则退出.输出最小区间长度[l,r]即可. #include<iostream> #include<alg…

题目 传送门:QWQ 分析 一开始没看到都是正整数根本不会做...... 看到了就是水题了.(但还是sb WA了一发) 尺取法搞一搞 代码 #include <bits/stdc++.h> using namespace std; + ; int A[maxn], B[maxn]; int main(){ int n,s; ){ ;i<=n;i++) scanf("%d",&A[i]); B[]=; ;i<=n;i++) B[i]=B[i-]+A[i];…

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]+--…

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,…

poj3061 Subsequence 题目链接: http://poj.org/problem?id=3061 挑战P146.题意:给定长度为n的数列整数a0,a1,...,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,如果解不存在,则输出零.$10<n<10^5,0<a_i<=10^4,S<10^8$; 思路:尺取法,设起始下标s,截止下标e,和为sum,初始时s=e=0;若sum<S,将sum增加a(e),并将e加1,若sum>=S,更…

传送门 NanoApe Loves Sequence Ⅱ Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)Total Submission(s): 1585    Accepted Submission(s): 688 Description NanoApe, the Retired Dog, has returned back to prepare for for the…

描述 http://poj.org/problem?id=2566 给出一个整数序列,并给出非负整数t,求数列中连续区间和的绝对值最接近k的区间左右端点以及这个区间和的绝对值. Bound Found Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 2592   Accepted: 789   Special Judge Description Signals of most probably extra-terrestr…

描述 http://poj.org/problem?id=3061 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,如果解不存在输出0. Subsequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11604   Accepted: 4844 Description A sequence of N positive integers (10 < N < 100 000), eac…

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…

Vasya and String time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only.…

题目链接:NanoApe Loves Sequence Ⅱ Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)Total Submission(s): 339    Accepted Submission(s): 165 Problem Description NanoApe, the Retired Dog, has returned back to prepare for…

1553: Good subsequence Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: 794  Solved: 287[Submit][Status][Web Board] Description Give you a sequence of n numbers, and a number k you should find the max length of Good subsequence. Good subsequence is a c…