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Divide the Sequence 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5783 Description Alice has a sequence A, She wants to split A into as much as possible continuous subsequences, satisfying that for each subsequence, every its prefix sum is not smal…

Divide the Sequence 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5783 Description Alice has a sequence A, She wants to split A into as much as possible continuous subsequences, satisfying that for each subsequence, every its prefix sum is not smal…

Divide the Sequence Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 649    Accepted Submission(s): 331 Problem Description Alice has a sequence A, She wants to split A into as much as possible c…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5783 题目大意: 把一个N个数的数列拆成若干段,保证每一段的前缀和都非负,求最多能拆成多少段. 题目思路: [贪心] 一开始题目看错了看成每一段内和非负..DPWA了好久. 默认答案是n,从后往前找负数,找到一个负数就一直把它往前合并直到和值非负,这样这个区间的前缀和就一定非负,扣除合并的区间大小即可. // //by coolxxx //#include<bits/stdc++.h> #inc…

Description Alice has a sequence A, She wants to split A into as much as possible continuous subsequences, satisfying that for each subsequence, every its prefix sum is not small than 0. Input The input consists of multiple test cases. Each test case…

HDU 5063 Operation the Sequence 题目链接 把操作存下来.因为仅仅有50个操作,所以每次把操作逆回去执行一遍,就能求出在原来的数列中的位置.输出就可以 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; const int N = 100005; const ll MOD = 1000…

Divide the Sequence 题意: 给你一个序列A,问你最多能够分成多少个连续子序列,使得每个子序列的所有前缀和均不小于0 题解: 这题是比赛时候的水题,但我比的时候也就做出这一题, = = 首先我想的是把他们前缀和求出来,之后试了下样例,一点鸟用都没有,那正着不行就倒着试下呗,之后发现这样有用,我先想到的思路是求后缀和,只要>=0就ans++,但那时队友举出了反例,比如2 1 -3 3的时候应该是2,我的算法就是4,那接着马上就能想到如果后缀和大于0,那么就要把他赋为0,之后举了几…

// 判断相同区间(lazy) 多校8 HDU 5828 Rikka with Sequence // 题意:三种操作,1增加值,2开根,3求和 // 思路:这题与HDU 4027 和HDU 5634 差不多 // 注意开根号的话,遇到极差等于1的,开根号以后有可能还是差1.如 // 2 3 2 3... // 8 9 8 9... // 2 3 2 3... // 8 9 8 9... // 剩下就是遇到区间相等的话,就直接开根号不往下传 #include <bits/stdc++.h> u…

题目链接:hdu 4983 Wow! Such Sequence! 题目大意:就是三种操作 1 k d, 改动k的为值添加d 2 l r, 查询l到r的区间和 3 l r. 间l到r区间上的所以数变成近期的斐波那契数,相等的话取向下取. 解题思路:线段树.对于每一个节点新增一个bool表示该节点下面的位置是否都是斐波那契数. #include <cstdio> #include <cstring> #include <cstdlib> #include <algo…