17997 Simple Counting

时间限制:2000MS  内存限制:65535K
提交次数:0 通过次数:0

题型: 编程题   语言: 不限定

Description

Ly is crazy about counting . Recently , he got a simple problem , but he had to learn Gaoshu these days .So , he turns to you for help .
You are given a sequence A with n positive integers numbered from 1 to n , and then expected to answer Q queries .
Each queries contains an interval [L,R] , you should find the number of index i which satisfies :
{i | Ai mod (i-L+1) = 0 , L <= i <= R }

where Ai mod (i-L+1) = 0 means that Ai can be divided by (i-L+1) .

输入格式

The first line of the input is an integer T , indicates the number of test cases .
Then T cases followed. For each test case :
The first line contains two integers n, Q .
The second line contains n positive integers A1, A2, …. An .
The next Q line , each line contains two integers L, R. Data range :
1<= T <= 20
1 <= n, Q <= 20000
1<= Ai <= 50000
1<= L <= R <= n

输出格式

For each query, output a single line with an integer indicates the answer expected .

输入样例

2
5 2
1 2 3 4 5
1 5
3 5 6 3
10 7 3 6 24 11
1 3
2 5
5 6

输出样例

5
2
2
3
1

提示

Huge input, scanf is preferred for C/C++.

During first sample ,
for the first query ,A1 mod 1 = 0 , A2 mod 2 = 0 , A3 mod 3 = 0 , A4 mod 4 = 0 ,A5 mod 5 = 0 , so the answer is 5 ;
for the second query , A3 mod 1 = 0 , A4 mod 2 = 0 , A5 mod 3 != 0 , so the answer is 2 .

给定n个数字和m次询问,每次给定区间[L, R],然后问a[L] % 1 == 0?  a[L + 1] % 2 == 0?,统计答案。

思路:考虑固定左端点L, 就是,对于每一个数字a[i],首先我们已经知道他在数组里的位置是i。

那么如果k是他的约数的话。如果真有询问问到它是否%k==0?,那么这个时候它应该从哪里开始这段区间的询问呢?

应该要在[i - k + 1, ]处吧,因为i - L + 1 = k。得到L等于这个。

然后就可以用一个vector[L]来保存,以L为开始的区间询问,那些位置会是得到ans的。然后二分小于等于R的个数就可以了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL; #include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = + ;
vector<int>pos[maxn];
void work() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = ; i <= n; ++i) {
pos[i].clear();
}
for (int i = ; i <= n; ++i) {
int x;
scanf("%d", &x);
int end = (int)sqrt(x * 1.0);
for (int j = ; j <= end; ++j) {
if (x % j == ) {
int L = i - j + ;
if (L >= ) {
pos[L].push_back(i);
} else break;
if (x / j == j) continue;
L = i - (x / j) + ;
if (L >= ) {
pos[L].push_back(i);
}
}
}
}
for (int i = ; i <= n; ++i) {
sort(pos[i].begin(), pos[i].end());
}
for (int i = ; i <= m; ++i) {
int L, R;
scanf("%d%d", &L, &R);
int ans = upper_bound(pos[L].begin(), pos[L].end(), R) - pos[L].begin();
printf("%d\n", ans);
}
} int main() {
#ifdef local
freopen("data.txt","r",stdin);
#endif
int t;
scanf("%d", &t);
while (t--) work();
return ;
}

17997 Simple Counting 数学的更多相关文章

  1. Codeforces 911D. Inversion Counting (数学、思维)

    题目链接:Inversion Counting 题意: 定义数列{ai|i=1,2,...,n}的逆序对如下:对于所有的1≤j<i≤n,若ai<aj,则<i,j>为一个逆序对. ...

  2. TOJ 1258 Very Simple Counting

    Description Let f(n) be the number of factors of integer n. Your task is to count the number of i(1 ...

  3. ACM学习历程—HDU5490 Simple Matrix (数学 && 逆元 && 快速幂) (2015合肥网赛07)

    Problem Description As we know, sequence in the form of an=a1+(n−1)d is called arithmetic progressio ...

  4. 「10.8」simple「数学」·walk「树上直径」

    A. Simple 本来以为很难,考场瞎推了推好像会了...... 想起小凯的诱惑,迷?? 首先$n$,$m$,$q$同除$gcd(n,m)$,显然$q$以内的数假如不是$gcd$的倍数,那么一定不能 ...

  5. UVA 11401 - Triangle CountingTriangle Counting 数学

    You are given n rods of length 1,2, . . . , n. You have to pick any 3 of them and build a triangle. ...

  6. [CSP-S模拟测试]:Simple(数学)

    题目描述 对于给定正整数$n,m$,我们称正整数$c$为好的,当且仅当存在非负整数$x,y$,使得$n\times x+m\times y=c$. 现在给出多组数据,对于每组数据,给定$n,m,q$, ...

  7. zoj 3286 Very Simple Counting---统计[1,N]相同因子个数

    Very Simple Counting Time Limit: 1 Second      Memory Limit: 32768 KB Let f(n) be the number of fact ...

  8. 13 Stream Processing Patterns for building Streaming and Realtime Applications

    原文:https://iwringer.wordpress.com/2015/08/03/patterns-for-streaming-realtime-analytics/ Introduction ...

  9. [C5] Andrew Ng - Structuring Machine Learning Projects

    About this Course You will learn how to build a successful machine learning project. If you aspire t ...

随机推荐

  1. hdu 1236 排名(排序)

    题意:按成绩排序 思路:排序 #include<iostream> #include<stdio.h> #include<string.h> #include< ...

  2. HihoCoder 1590 : 紧张的会议室(区间最大+离散化)

    时间限制:20000ms 单点时限:2000ms 内存限制:256MB 描述 小Hi的公司最近员工增长迅速,同时大大小小的会议也越来越多:导致公司内的M间会议室非常紧张. 现在小Hi知道公司目前有N个 ...

  3. bzoj 2006 [NOI2010]超级钢琴——ST表+堆

    题目:https://www.lydsy.com/JudgeOnline/problem.php?id=2006 每个右端点的左端点在一个区间内:用堆记录端点位置.可选区间,按价值排序:拿出一个后也许 ...

  4. javascript 前段MVVM 框架

    http://www.likebin.net/meteorlist.html http://www.cnblogs.com/sskyy/p/3197917.html

  5. Codeforces 1108F MST Unification MST + LCA

    Codeforces 1108F MST + LCA F. MST Unification Description: You are given an undirected weighted conn ...

  6. (转)data Table的用法大全

    jqyery dataTable 基本用法 一:官方网站:[http://www.datatables.net/] 二:基本使用:[http://www.guoxk.com/node/jquery-d ...

  7. python数据分析笔记中panda(1)

    1 例子1 from pandas import read_csv; df = read_csv('H://pythonCode//4.1//1.csv') df 截图 1.1 修改表的内容编码 df ...

  8. AttributeCollection.Add(String, String) Method

    <%@ Page Language="C#" %> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Trans ...

  9. PLSQL配置教程

    下载一个PLSQL安装包 网盘下载地址↓ 链接: https://pan.baidu.com/s/1q-uwAfeLOPxzBBx6V1pYLg 提取码: hei9 PLSQL文件夹一定要放在D:\i ...

  10. LeetCode: 620 Not Boring Movies(easy)

    题目: X city opened a new cinema, many people would like to go to this cinema. The cinema also gives o ...