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

2
2
1
1
0

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
");
                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
",R-L+1);

        }
    }

    return 0;
}

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