codeforces 594D题解

D. REQ

time limit per test3 seconds
memory limit per test256 megabytes
input standard input
output standard output
Today on a math lesson the teacher told Vovochka that the Euler function of a positive integer φ(n) is an arithmetic function that counts the positive integers less than or equal to n that are relatively prime to n. The number 1 is coprime to all the positive integers and φ(1) = 1.

Now the teacher gave Vovochka an array of n positive integers a1, a2, …, an and a task to process q queries li ri — to calculate and print modulo 109 + 7. As it is too hard for a second grade school student, you’ve decided to help Vovochka.

Input

The first line of the input contains number n (1 ≤ n ≤ 200 000) — the length of the array given to Vovochka. The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 106).

The third line contains integer q (1 ≤ q ≤ 200 000) — the number of queries. Next q lines contain the queries, one per line. Each query is defined by the boundaries of the segment li and ri (1 ≤ li ≤ ri ≤ n).

Output

Print q numbers — the value of the Euler function for each query, calculated modulo 109 + 7.

Sample test(s)

input
10
1 2 3 4 5 6 7 8 9 10
7
1 1
3 8
5 6
4 8
8 10
7 9
7 10
output
1
4608
8
1536
192
144
1152
input
7
24 63 13 52 6 10 1
6
3 5
4 7
1 7
2 4
3 6
2 6
output
1248
768
12939264
11232
9984
539136

Note

In the second sample the values are calculated like that:

φ(13·52·6) = φ(4056) = 1248
φ(52·6·10·1) = φ(3120) = 768
φ(24·63·13·52·6·10·1) = φ(61326720) = 12939264
φ(63·13·52) = φ(42588) = 11232
φ(13·52·6·10) = φ(40560) = 9984
φ(63·13·52·6·10) = φ(2555280) = 539136

算法

• 离线处理每一组询问
• 列出欧拉函数的公式 观察每一项素因子对最后答案的贡献
• 用树状数组维护前缀答案

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#define MAXN 200011
#define limit 1000011
int                                 n,a[MAXN],minp[limit],b[limit],v,prime[limit],t[MAXN],m,ans[MAXN];
typedef long long ll;
const ll mod=1e9+7;
using namespace std;
struct Node
{
    int l,r,num;
}q[MAXN];
inline bool cmp(Node a,Node b)
{
    return a.r<b.r;
}
inline void predeal()
{
    int tot=0;
    for (int i=2;i<=v;i++)
    {
        if(!minp[i])prime[tot++]=i,minp[i]=i;
        for (int j=0;j<tot;j++)
        {
            if(prime[j]*i>v)break;
            minp[prime[j]*i]=prime[j];
            if(i%prime[j]==0)break;
        }
    }
    for (int i=0;i<=n;i++)
        t[i]=1;
}
inline int pow(int a,ll b)
{
    int r=1,base=a;
    while(b)
    {
        if(b&1)r=(ll)r*base%mod;
        base=(ll)base*base%mod;
        b>>=1;
    }
    return r%mod;
}
inline void modify(int x,int z)
{
    for (int i=x;i<=n;i+=i&-i)
        t[i]=(ll)t[i]*z%mod;
}
inline int query(int x)
{
    int temp=1;
    for (int i=x;i;i-=i&-i)
        temp=(ll)temp*t[i]%mod;
    return temp;
}
inline void change(int p)
{
    int x=a[p];
    while(x!=1)
    {
        int now=minp[x];
        if(b[now])
        {
            modify(b[now],pow(now-1,mod-2));
            modify(b[now],now);
        }
        b[now]=p;
        modify(p,now-1);
        x/=now;
    }
}
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    v=*max_element(a+1,a+1+n);
    predeal();
    scanf("%d",&m);
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d",&q[i].l,&q[i].r);
        q[i].num=i;
    }
    sort(q+1,q+1+m,cmp);
    for (int i=1,now=1;i<=m;i++)
    {
        while(now<=q[i].r)change(now++);
        ans[q[i].num]=(ll)query(q[i].r)*pow(query(q[i].l-1),mod-2)%mod;
    }
    for (int i=1;i<=m;i++)
        printf("%d\n",ans[i]);
    return 0;
}