51nod 1642 interval Euler's Function & codeforce594d req

When you draw a persimmon, you will know that it is the formula for finding all the prime factors in the product multiplication interval (MA)/P (that is, the Euler's formula)

It seems that Mo team is very reliable, but time O (nsqrt (n) logn) is not porcelain

Or offline. determine the right endpoint ~ We insert the prime factor in the I range into the last position in the tree array, and scan the inverse element + find the prime factor O (nlog ^ 2n)

Note that we cannot use Trial Division when calculating the quality factor.

#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>using namespace std;typedef long long LL;const LL mod=1e9+7;int quick_pow(int A,int p){    int ret=1;    while(p!=0)    {        if(p%2==1)ret=(LL)ret*A%mod;        A=(LL)A*A%mod;p/=2;    }    return ret;}int inv(int A){return quick_pow(A,mod-2);}int pr,prime[1100000],pm[1100000];bool v[1100000];void get_prime(){    pr=0;    for(int i=2;i<=1001000;i++)    {        if(v[i]==false)prime[++pr]=i,pm[i]=i;        for(int j=1;j<=pr&&i*prime[j]<=1001000;j++)        {            v[i*prime[j]]=true;            pm[i*prime[j]]=min(pm[i],prime[j]);            if(i%prime[j]==0)break;        }    }}int n;LL s[210000];int lowbit(int x){return x&-x;}void change(int x,LL k){    while(x<=n)    {        s[x]=s[x]*k%mod;        x+=lowbit(x);    }}LL getsum(int x){    LL ret=1;    while(x>0)    {        ret=ret*s[x]%mod;        x-=lowbit(x);    }    return ret;}int a[210000];LL sm[210000];struct query{int l,r,id;}q[210000];int as[210000];bool cmp(query q1,query q2){return q1.r<q2.r;}int last[1100000];int main(){    get_prime();    scanf("%d",&n);    sm[0]=1;    for(int i=1;i<=n;i++)        scanf("%d",&a[i]), sm[i]=sm[i-1]*a[i]%mod;    int Q;    scanf("%d",&Q);    for(int i=1;i<=Q;i++)        scanf("%d%d",&q[i].l,&q[i].r), q[i].id=i;    sort(q+1,q+Q+1,cmp);        int j=1;    memset(last,0,sizeof(last));    for(int i=1;i<=n;i++)s[i]=1;    for(int i=1;i<=n;i++)    {        int d=a[i];        while(d>1)        {            int p=pm[d];LL c=(LL)(p-1)*inv(p)%mod;            if(last[p]>0)change(last[p],inv(c));            last[p]=i;            change(last[p],c);            while(d%p==0)d/=p;        }                while(j<=Q&&q[j].r==i)        {            as[q[j].id]=sm[q[j].r]*inv(sm[q[j].l-1])%mod*getsum(q[j].r)%mod*inv(getsum(q[j].l-1))%mod;            j++;        }    }        for(int i=1;i<=Q;i++)printf("%d\n",as[i]);    return 0;}

 

51nod 1642 interval Euler's Function & codeforce594d req