Minimal Subarray Length Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVALive. Original ID: 660964-bit integer IO format: %lld      Java class name: Main     You are given an integer sequence of length N and another value X…

题目链接 题意:给出n个数,求加和大于x的最短区间的区间长度. 如果前i个数字和为y,那么如果前j数字的和小于等于y-x,那么i-j就是一种可能的情况,我们对于所有的i找出前面最大的j就可以了,因为数据量比较大,用树状数组来处理前n项的最大值,但是每个数字又可能比较大,所以先离散化处理一下. AC代码 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 typedef long long LL;…

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4620 You are given an integer sequence of length N and another value X. You have to find a contiguoussubsequence of the given sequence such t…

描述:给定n个整数元素,求出长度最小的一段连续元素,使得这段元素的和sum >= X. 对整个数组先求出sum[i],表示前i个元素的和,然后依次求出以a[i]为起点的,总和>= X的最小长度, 每次考虑新元素a[i]时,将a[i]加入数组, pa[-q] = mp(sum[i], i),这样pa[q---.p]形成一段总和递增的序列,下标也是逐渐增大. 最后利用lower_bound函数求出大于或等于sum[i-1]+X的最左边的元素,求出以i为起点最短长度. 最后想了一下如果是求>=…

题意:给定长度为N的数组,求一段连续的元素之和大于等于K,并且让这段元素的长度最小,输出最小长度即可,若不存在这样的元素集合,则输出-1 题目链接:UVAlive 6609 做法:做一个前缀和prefix,然后再作一个维护前缀和最大值数组Max,枚举所有可能的起始点i,在Max上二分末尾位置r,由于Max维护的是前缀和的最大值,因此具有单调性,可以进行二分,似乎还有其他O(n)的做法,有空去膜一下 代码: #include <stdio.h> #include <bits/stdc++.…

10020 Given several segments of line (int the X axis) with coordinates [Li, Ri]. You are to choose the minimalamount of them, such they would completely cover the segment [0, M].InputThe first line is the number of test cases, followed by a blank lin…

可以说是区间覆盖问题的例题... Note: 区间包含+排序扫描: 要求覆盖区间[s, t]; 1.把各区间按照Left从小到大排序,如果区间1的起点大于s,则无解(因为其他区间的左起点更大):否则选择起点在s的最长区间; 2.选择区间[li, ri]后,新的起点应更新为ri,并且忽略所有区间在ri之前的部分:  Minimal coverage  The Problem Given several segments of line (int the X axis) with coordinat…

Given several segments of line (int the X axis) with coordinates [Li, Ri]. You are to choose the minimal amount of them, such they would completely cover the segment [0, M].InputThe first line is the number of test cases, followed by a blank line.Eac…

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=961 贪心,排序,对左端点贪,找最大右端点. #include <cstdio> #include <cstring> #include <algorithm> #define maxn 6000000 using namespace std; int t,…

 Minimal coverage  The Problem Given several segments of line (int the X axis) with coordinates [Li,Ri]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M]. The Input The first line is the number of test…