17997 Simple Counting Math

17997 Simple Counting

Time limit: 2000MS memory limit: 65535K
Number of submissions: 0 Number of Passes: 0

Question types: programming language: not Limited

Description

Ly is crazy about counting. Recently, he got a simple problem, but he had to learn gaoshu these days. So, he turns to your for help. You is given a sequence a with n positive integers numbered from 1 to N, and then expected to answer Q queries. Each queries contains a 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 is divided by (i-l+1).

Input format

The first line of the input was an integer T, indicates the number of test cases. Then T cases followed. For each test case:the first line contains the integers n, Q. The second line contains n positive integers A1, A2, .... An. The next Q line, all line contains, integers L, r.data range:1<= T <= 201 <= N, q <= 200001<= Ai < = 500001<= L <= R <= N

Output format

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

Input sample

25 21 2 3 4 51 53 56 310 7 3 6 24 111 32 55 6

Output sample

52231

Tips

Huge input, scanf is preferred for 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 = 0A3 MoD 1 = 0, A4 mod 2 = 0, A5 mod 3 ! = 0, so the answer is 2.

Given n numbers and M queries, each given interval [L, R], then ask a[l]% 1 = = 0? A[l + 1]% 2 = = 0?, statistical answer.

Idea: Consider fixed left endpoint L, that is, for each number a[i], first we already know that his position in the array is I.

Well, if K is his approximate. If you really ask if it is%k==0? , where should it start asking about this interval?

It should be in [I-k + 1,], because i-l + 1 = k. Get L equals this.

Then you can use a vector[l] to save, with L as the beginning of the interval to ask, those positions will be the ANS. Then the two points are less than or equal to the number of R.

#include <cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#defineIOS Ios::sync_with_stdio (False)using namespacestd;#defineINF (0X3F3F3F3F)typedefLong Long intLL; #include<iostream>#include<sstream>#include<vector>#include<Set>#include<map>#include<queue>#include<string>Const intMAXN =20000+ -; Vector<int>POS[MAXN];voidWork () {intN, M; scanf ("%d%d", &n, &m);  for(inti =1; I <= N; ++i) {pos[i].clear (); }     for(inti =1; I <= N; ++i) {intx; scanf ("%d", &x); intEnd = (int) sqrt (x *1.0);  for(intj =1; J <= End; ++j) {if(x% J = =0) {                intL = I-j +1; if(L >=1) {pos[l].push_back (i); } Else  Break; if(x/j = = j)Continue; L= i-(x/j) +1; if(L >=1) {pos[l].push_back (i); }            }        }    }     for(inti =1; I <= N; ++i) {sort (Pos[i].begin (), Pos[i].end ()); }     for(inti =1; I <= m; ++i) {intL, R; scanf ("%d%d", &l, &R); intAns = Upper_bound (Pos[l].begin (), Pos[l].end (), R)-Pos[l].begin (); printf ("%d\n", ans); }}intMain () {#ifdef local freopen ("Data.txt","R", stdin);#endif    intT; scanf ("%d", &t);  while(t--) work (); return 0;}

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17997 Simple Counting Math