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 1000011int                                 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;}