HDU 5288 OO’s Sequence http://acm.hdu.edu.cn/showproblem.php?pid=5288 OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there’s no j(l<=j<=r,j<>i) satisfy a i mod a j=0,now OO want to k…

题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=5288 题面: OO's Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 985    Accepted Submission(s): 375 Problem Description OO has got a arra…

OO's Sequence 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5288 Description OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO wan…

OO’s Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1549    Accepted Submission(s): 559 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the nu…

OO's Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1751    Accepted Submission(s): 632 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the n…

OO's Sequence                                                          Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)                                                                                             T…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5288 题意:在闭区间[l,r]内有一个数a[i],a[i]不能整除 除去自身以外的其他的数,f(l,r)表示在这区间内a[i]这样的数的个数,,现给你n个数,求所有区间的f(l,r)的和. 思路:对于每个数a[i]求出他的左右侧最靠近他的且是他的因子的位置L.R,并记录,那么对于每个数a[i]都有了他的L,R,而对于每个a[i]在f(l,r)有价值的次数之和就是(i-L+1)*(R-i+1) 代码:…

Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO want to know ∑i=1n∑j=inf(i,j) mod (109+7).   Input There are m…

Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO want to know ∑i=1n∑j=inf(i,j) mod (+). Input There are multipl…

题意:给一个序列,函数f(l, r)表示在[l, r]区间内有多少数字不是其他数字的倍数,求所有区间的f(l, r)之和. 解法:第一次打多校……心里还有点小激动……然而一道签到题做了俩点……呜呜呜……今天的题还算简单……明天就更难了……写个题解纪念一下多校…… 对于序列中的每一个数,要找到从它的位置起向左右找最远连续不能被它整除的数的位置设为l和r,这个数的位置为pos,答案就是(pos - l + 1) * (r - pos + 1),只要分析一下样例就可以得到这个式子……然后为了找到l和r…

题意: 给你一个序列, 有一个函数 F(L,R) 其中 ai 均不能 被 aL - aR整除的  函数值是这个ai个数 思路 : 反过来求 满足这样的条件的 ai 的区间,然后求和 #include<iostream> #include<vector> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef __int64 LL; ;…

题目传送门 /* 定义两个数组,l[i]和r[i]表示第i个数左侧右侧接近它且值是a[i]因子的位置, 第i个数被选择后贡献的值是(r[i]-i)*(i-l[i]),每个数都枚举它的因子,更新l[i], r[i],复杂度O(n*sqrt(a[i])) */ #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #include <map> using…

OO's Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 955    Accepted Submission(s): 358 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the nu…

预处理出每一个数字的左右两边能够整除它的近期的数的位置 OO's Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1880    Accepted Submission(s): 672 Problem Description OO has got a array A of size n ,defined a fun…

OO's Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 2643    Accepted Submission(s): 925 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the n…

题意: 给你一个序列, 有一个函数 F(L,R) 其中 ai 均不能 被 aL - aR整除的  函数值是这个ai个数 思路 : 反过来求 满足这样的条件的 ai 的区间,然后求和 #include<iostream> #include<vector> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef __int64 LL; ;…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5288 //*************头文件区************* #include<iostream> #include<cstdio> #include<vector> #define N 100010 #define P 1000000007 using namespace std; int n,tmp,i,j; int r[N], l[N], q[N], a[N…

二分寻找对于指定pos的最左因数点和最右因数点. /* 5288 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <deque> #include <algorithm> #includ…

Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO want to know     Input There are multiple test cases. Please p…

OO’s Sequence Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=5288 Mean: 给定一个数列,让你求所有区间上满足Ai%Aj!=0(Ai!=Aj)的Ai的个数之和. analyse: 对于Ai,如果我们知道最靠近Ai且能够整除Ai的数的下标l和r,那么Ai对答案的贡献就是(r-i)*(i-l).剩下的就是怎样去求每个Ai的l和r了. 首先我们预处理出:对于每个i,能够被1~i整除的数,用链表存起来. 那么对于输…

pid=5288">http://acm.hdu.edu.cn/showproblem.php?pid=5288 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now…

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// 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…

题目链接: Hdu 5496 Beauty of Sequence 题目描述: 一个整数序列,除去连续的相同数字(保留一个)后,序列的和成为完美序列和.问:一个整数序列的所有子序列的完美序列和? 解题思路: 考虑位于i位置数字x的贡献值,假设x是子序列中连续相同数字的第一个,那么x对于i后面的数有2(n-i)个贡献值,对前面的数,要么不选取,要么选取结尾不为x的方案数目. #include <map> #include <cstdio> #include <cstring&g…

Hdu 5806 NanoApe Loves Sequence Ⅱ(双指针) Hdu 5806 题意:给出一个数组,求区间第k大的数大于等于m的区间个数 #include<queue> #include<cmath> #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define ll l…

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…

OO’s Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 312    Accepted Submission(s): 107 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the num…

OO's Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 42    Accepted Submission(s): 21 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the numb…

HDU - 1711 A - Number Sequence   Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... ,…