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

6 512 8 17 15 90 281 4 32 6 51 1 23 5 171 6 999983

Sample Output

22110

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，然后对于数组每个数分解质因子，将相同的质因子作为下标，存包含这些质因子的那些数的坐标，然后可以直接查询范围

#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;}