CSU1661: Query Mutiple

Description

One day，Little-Y saw many numbers standing in a row. A question suddenly appeared in her mind, ”From the L-th number to the R-th number, how many of them is a mutiple of P ? (P is a prime number) , and how quickly can I settle this problem ? ”

Input

Mutiple test cases. Please process till the end of file.

For each test case：

The first line contains two positive integers n and q (1<=n,q<=10^5), which means there are n integers standing in a row and q queries followed.

The second line contains n positive integers, a1,a2,a3,…,an (no more than 10^6) . Here, ai means the i-th integer in the row.

The next are q queries, each of which takes one line. For each query, there are three integers L,R,P (1<=L<=R<=n, 1<=P<=10^6, P is gurantteed to be a prime number). Their meanings have been mentioned in the discription.

Output

For each query, output the answer in one line.

Sample Input

6 5

12 8 17 15 90 28

1 4 3

2 6 5

1 1 2

3 5 17

1 6 999983

Sample Output

HINT

Source

题意：

给出一个数组

然后m个询问，每次询问l,r,p,询问数组l~r区间内有几个p的倍数

思路：

首先打出素数表。然后对于数组每一个数分解质因子。将同样的质因子作为下标。存包括这些质因子的那些数的坐标，然后能够直接查询范围

#include <iostream>

#include <stdio.h>

#include <string.h>

#include <string>

#include <stack>

#include <queue>

#include <map>

#include <set>

#include <vector>

#include <math.h>

#include <bitset>

#include <list>

#include <algorithm>

#include <climits>

using namespace std;

#define lson 2*i

#define rson 2*i+1

#define LS l,mid,lson

#define RS mid+1,r,rson

#define UP(i,x,y) for(i=x;i<=y;i++)

#define DOWN(i,x,y) for(i=x;i>=y;i--)

#define MEM(a,x) memset(a,x,sizeof(a))

#define W(a) while(a)

#define gcd(a,b) __gcd(a,b)

#define LL long long

#define N 1000000

#define INF 0x3f3f3f3f

#define EXP 1e-8

#define lowbit(x) (x&-x)

const int mod = 1e9+7;

vector<int> prime[N+5];

bool vis[N+5];

int pos[N+5],tot;

void set()

{

    int i,j;

    memset(vis,1,sizeof(vis));

    for(i = 2; i<=N; i++)

    {

        if(vis[i])

        {

            if(N/i<i)

                break;

        }

        for(j = (LL)i+i; j<=N; j+=i)

            vis[j] = 0;

    }

    tot = 1;

    for(i = 2; i<=N; i++)

        if(vis[i])

            pos[i] = tot++;

}

int main()

{

    set();

    int i,j,n,m,x,y,l,r,L,R;

    while(~scanf("%d%d",&n,&m))

    {

        for(i = 1; i<tot; i++)

            prime[i].clear();

        for(i = 1; i<=n; i++)

        {

            scanf("%d",&x);

            for(j = 2; j*j<=x; j++)

            {

                if(x%j!=0)

                    continue;

                prime[pos[j]].push_back(i);

                while(x%j==0)

                    x/=j;

            }

            if(x>1)

                prime[pos[x]].push_back(i);

        }

        for(i = 1; i<tot; i++)

            sort(prime[i].begin(),prime[i].end());

        while(m--)

        {

            scanf("%d%d%d",&l,&r,&x);

            y = pos[x];

            int len = prime[y].size();

            if(len==0||l>prime[y][len-1]||r<prime[y][0])

            {

                printf("0\n");

                continue;

            }

            L = lower_bound(prime[y].begin(),prime[y].end(),l)-prime[y].begin();

            R  = lower_bound(prime[y].begin(),prime[y].end(),r)-prime[y].begin();

            if(R==len||prime[y][R]>r)

                R--;

            printf("%d\n",R-L+1);

        }

    }

    return 0;

}