Gym 101102J---Divisible Numbers(反推技巧题)

题目链接

http://codeforces.com/gym/101102/problem/J

Description

standard input/output

You are given an array A of integers of size N, and Q queries. For each query, you will be given a set of distinct integers S and two integers L and R that represent a range in the array. Your task is to count how many numbers in the given range are divisible by at least one number from the set.

Input

The first line of input contains a single integer T, the number of test cases.

The first line of each test case contains two integers, N and Q (1 ≤ N, Q ≤ 10⁵), the size of the array and the number of queries, respectively.

The next line contains N space-separated integers, the values of the array A (1 ≤ A_i ≤ 10⁹).

Each of the next Q lines contain the description of one query in the form:

LRS

Where L and R (1 ≤ L ≤ R ≤ N) represent the range, and S is an integer between 1 and 1023 (inclusive) and represents the set; consider the binary representation of the number S, if the i^th bit (1-based) is 1, then the number i belongs to the set. Since S is less than1024, the values in the set are between 1 and 10.

For example: if S is equal to 6, the binary representation of 6 is 110, and this means the values in the set are 2 and 3.

The input was given in this way to reduce the size of the input file.

Output

Print the answer for each query on a single line.

Sample Input

Input

Output

1
3

题意：输入n，q表示由n个数构成的一个序列，q次询问，每次询问输入l r s s表示一个集合，s的二进制中的1的位置，如：6（10）=110（2） 对应集合{2,3} 现在求区间l~r中能被集合中的（任意一个）数整除的数的个数；

思路：因为s取值范围为1~1023 所以输入n个数时直接求出能被哪些s（对应的集合数）整除，然后用前缀和表示出来，那么每次查询时就是0（1）的了；

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
const int MAXN = 1e5+5;
int sum[512][MAXN];
template <class T>
inline void Scan(T &ret)
{
    char c=getchar();
    while(c<'0'||c>'9')
        c=getchar();
    ret=c-'0';
    while(c=getchar(),c>='0'&&c<='9')
        ret=ret*10+(c-'0');
}

int main()
{
    int T;
    Scan(T);
    while(T--)
    {
        int n,q,x,s;
        Scan(n); Scan(q);
        ///memset(sum,0,sizeof(sum));  加上超时！！！
        for(int i=1;i<=n;i++)
        {
            s=0;
            Scan(x);
            for(int j=2;j<=10;j++)
                if(x%j==0) s=s|(1<<(j-2));
                
            for(int j=0;j<512;j++)
                sum[j][i]=sum[j][i-1]+((s&j)?1:0);
        }
        while(q--)
        {
            int l,r;
            Scan(l); Scan(r); Scan(x);
            if(x&1) printf("%d
",r-l+1);
            else    printf("%d
",sum[x>>1][r]-sum[x>>1][l-1]);
        }
    }
    return 0;
}