Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 3636    Accepted Submission(s): 1612 Problem Description In a galaxy far, far away, there are two integer sequence a and b of l…

原题地址 Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 887    Accepted Submission(s): 336 Problem Description In a galaxy far, far away, there are two integer sequence a and b o…

题意:给定一个初始数组b和一个初始值全部为0的数组a,每次操作可以在给定的区间(l,r)内让a[i](l=<i<=r)加一,或者查询区间区间(l,r)中a[i]/b[i](l=<i<=r)(取整)的和.可以知道,$\sum_{\frac{a_i}{b_i}}\le nlogn$,所以我们只要暴力找到需要修改的位置修改即可..代码: #include<iostream> #include<cstdio> #include<cstring> #def…

构造一个序列B[i]=-b[i],建一颗线段树,维护区间max, 每次区间加后再询问该区间最大值,如果为0就在树状数组中对应的值+1(该操作可能进行多次) 答案在树状数组中找 其实只用一颗线段树也是可以的 Code #include <cstdio> #include <algorithm> #include <cstring> #define mst(a) memset(a,0,sizeof(a)) #define N 100010 using namespace s…

题意 题目链接 Sol 这题关键是注意到题目中的\(b\)是个排列 那么最终的答案最多是\(nlogn\)(调和级数) 设\(d_i\)表示\(i\)号节点还需要加\(d_i\)次才能产生\(1\)的贡献 用线段树维护每个节点里\(d_i\)的最小值,每次当\(d_i - 1= 0\)的时候往下递归即可 时间复杂度:\(O(nlog^2 n)\) 多组数据记得清空lazy标记啊qwq.... #include<bits/stdc++.h> using namespace std; const…

目录 Catalog Solution: (有任何问题欢迎留言或私聊 && 欢迎交流讨论哦 Catalog Problem:Portal传送门  原题目描述在最下面. Solution:  每个节点用一个变量储存它所覆盖区间最少需要加多少次答案能加1.  如果这次更新不能使答案加1,则更新到lazy标记就可以了,如果能让答案加1,就更新到叶子节点,具体看代码. AC_Code: #include<cstdio> #include<iostream> #include…

链接:HDU-6315:Naive Operations 题意: In a galaxy far, far away, there are two integer sequence a and b of length n.b is a static permutation of 1 to n. Initially a is filled with zeroes.There are two kind of operations:1. add l r: add one for al,al+1...a…

题目: http://acm.hdu.edu.cn/showproblem.php?pid=6315 Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others) Problem Description In a galaxy far, far away, there are two integer sequence a and b of length…

Naive Operations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 502768/502768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description In a galaxy far, far away, there are two integer sequence a and b of length…

题目大意 题目链接Naive Operations 题目大意: 区间加1(在a数组中) 区间求ai/bi的和 ai初值全部为0,bi给出,且为n的排列,多组数据(<=5),n,q<=1e5 axmorz 思路 因为是整除,所以一次加法可以对ans 没有影响 当ai是bi的倍数,对ans会有贡献 所以我们维护一个sum,初值为bi(只对于线段树的叶子节点有用) 当区间+1的时候 我们对sum-1 当sum=0的时候(倍数) ans++,sum=bi 然后再维护一个区间最小值 当区间内的mi>…