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&…

1029. 可被 5 整除的二进制前缀  显示英文描述 我的提交返回竞赛   用户通过次数467 用户尝试次数662 通过次数477 提交次数1964 题目难度Easy 给定由若干 0 和 1 组成的数组 A.我们定义 N_i:从 A[0] 到 A[i] 的第 i 个子数组被解释为一个二进制数(从最高有效位到最低有效位). 返回布尔值列表 answer,只有当 N_i 可以被 5 整除时,答案 answer[i] 为 true,否则为 false. 示例 1: 输入:[0,1,1] 输出:[tr…

题目链接:http://poj.org/problem?id=3061 题意:给一个长为n的数列和整数s,求一个连续的子序列,使得这个子序列长度最短并且不小于这个整数s. 统计[1~i]的子序列和sum(i),(sum(0)=0).然后求一个区间[i,j]的和即为sum(j)-sum(i-1) (i > 0). 由于给定序列没有负数,因此sum是个严格不减的序列. 转换成一个求最大值最小的问题,可以二分枚举序列长度,在前缀和上计算子序列[i-1,i+m-1]的和.如果存在一个满足子序列和≥s的,…

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…

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…

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: 16520   Accepted: 7008 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…

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,…

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…

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…