题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5875 Function Time Limit: 7000/3500 MS (Java/Others)Memory Limit: 262144/262144 K (Java/Others) 问题描述 The shorter, the simpler. With this problem, you should be convinced of this truth. You are given an…

Function Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 2427    Accepted Submission(s): 858 Problem Description The shorter, the simpler. With this problem, you should be convinced of this t…

Function Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 976    Accepted Submission(s): 375 Problem Description The shorter, the simpler. With this problem, you should be convinced of this tru…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5875 [题目大意] 给出一个数列,同时给出多个询问,每个询问给出一个区间,要求算出区间从左边开始不断对下一个数取模之后的结果. [题解] 考虑取模的递减性质,最多取模logn次,因此如何快速找出下一个取模的位置是解决这道题的关键,首先利用ST表预处理区间最小值,之后我们每次二分取模起点到终点这个区段,每次优先查询左区间(类似于线段树上的路径查询),这样子就能logn找到取模的位置,然后直接取模就…

Function Problem Description   The shorter, the simpler. With this problem, you should be convinced of this truth.    You are given an array A of N postive integers, and M queries in the form (l,r). A function F(l,r) (1≤l≤r≤N) is defined as:F(l,r)={A…

Function Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1095    Accepted Submission(s): 420 Problem Description The shorter, the simpler. With this problem, you should be convinced of this tr…

Function Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1498    Accepted Submission(s): 553 Problem Description The shorter, the simpler. With this problem, you should be convinced of this t…

Function The shorter, the simpler. With this problem, you should be convinced of this truth.      You are given an array AA of NN postive integers, and MM queries in the form (l,r)(l,r). A function F(l,r) (1≤l≤r≤N)F(l,r) (1≤l≤r≤N) is defined as: F(l,…

题目链接 网络赛的水实在太深,这场居然没出线zzz,差了一点点,看到这道题的的时候就剩半个小时了.上面是官方的题意题解,打完了才知道暴力就可以过,暴力我们当时是想出来了的,如果稍稍再优化一下估计就过了zzz.去年有一场现场赛也是n=1000,n^3过了,看来关键时刻实在做不出来就得大胆暴力啊. #include <bits/stdc++.h> using namespace std; typedef long long ll; ; int a[maxn],nex[maxn]; int main…

很简单的一个题的,结果后台数据有误,自己又太傻卡了3个小时... 题意:给你一串数a再给你一些区间(lef,rig),求出a[lef]%a[lef+1]...%a[rig] 题解:我们可以发现数字a对数字b取模时:如果a<b,则等于原数,否则a会变小至少一半.就是说a最多成功取模(log2 a)次,所以我们只需要每次在区间内找到最前面一个小于等于a的值,接着更新a与区间左端点,直到没有值比a小或者区间取模完成. 我们可以使用线段树求出区间内小于某个值的最前一个位置,具体方法就是:父节点记录区间最…