牛客练习赛9 珂朵莉的约数 小数据暴力大数据莫队

题目链接

 

首先预处理质数和逆元。

我们发现，如果直接莫队，复杂度会很高，根据题解而言，

一个数的约数是[1000,1000000]之间的质数最多只有一个，

那么对于(1,1000]的质数用前缀和维护sum[i][j]代表前i个数第j个质数个数和。

那么对于每个a[i]处理处sum[i][j]大于1000的数每个a[i]最多只有一个存在b[i]里

使用莫队处理b[i]部分即可。

答案就是每个质数个数+1的乘积

 

 

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
//#pragma GCC optimize("Ofast")
//#pragma comment(linker, "/STACK:102400000,102400000")
//#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx) 
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <fstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e6+100;
const int maxx=1e7+10;
const double EPS=1e-8;
const double eps=1e-8;
const LL mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int n,m;
int a[maxn];
int b[maxn],c[maxn];
LL sum[maxn/10][205];
int pos[maxn];
struct node {
    int l,r,id;
}q[maxn];

LL xx;
LL ans[maxn];
LL inv[maxn];
LL cnt[maxn];
int tot,pr[maxn],vis[maxn];
void init()//绱犳暟琛? O(nsqrt(n))
{
    for(int i=2;i<maxn;i++)
    {
        if(!vis[i])pr[tot++]=i;
        for(int j=0;j<tot&&i*pr[j]<maxn;j++)
        {   
            vis[i*pr[j]]=1; 
            if(i%pr[j]==0)break;
         }
    }
    inv[0]=inv[1]=1;
    for(int i=2;i<=maxn;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
}

bool cmp(node a,node b){
    if(pos[a.l]==pos[b.l])
        return a.r<b.r;
    return a.l<b.l;
}

void add(int x){
    if(x==0) return ;
    xx=1ll*xx*inv[1+cnt[x]]%mod;
    cnt[x]++;
    xx=1ll*xx*(1+cnt[x])%mod;  
}

void del(int x){
    if(x==0) return ;
    xx=1ll*xx*inv[1+cnt[x]]%mod;
    cnt[x]--;
    xx=1ll*xx*(1+cnt[x])%mod;
}
void solve(){
    init();
    s_2(n,m);
    int sz=sqrt(n);
    FOR(1,n,i){
        pos[i]=i/sz;
        s_1(a[i]);
        int tmp=a[i];
        int g=0;
        int now=0;
        W(tmp!=1){
            if(!vis[tmp]){
                c[g++]=tmp;
                tmp=1;
            }
            else {
                W(tmp%pr[now]==0){
                    c[g++]=pr[now];
                    tmp=tmp/pr[now];
                }
                now++;
            }
        }
        FOr(0,g,j){
            if(c[j]>1000) b[i]=c[j];
            else {
                FOr(0,200,k){
                    if(pr[k]==c[j]){
                        sum[i][k]++;
                    }
                }
            }
        }
    }
    FOR(1,n,i)
        FOR(0,200,j)
            sum[i][j]=sum[i-1][j]+sum[i][j];
    
    FOR(1,m,i){
        s_2(q[i].l,q[i].r);
        q[i].id=i;
    }
    sort(q+1,q+1+m,cmp);
    xx=1;
    int l=1,r=0;
    FOR(1,m,i){
        W(r<q[i].r) r++,add(b[r]);
        W(r>q[i].r) del(b[r]),r--;
        W(l<q[i].l) del(b[l]),l++;
        W(l>q[i].l) l--,add(b[l]);
        LL gg=1;
        FOr(0,200,j){
            gg=1ll*gg*(sum[r][j]-sum[l-1][j]+1)%mod;
        }
        ans[q[i].id]=1ll*xx*gg%mod;
    }
    FOR(1,m,i) 
        print(ans[i]);
}
int main() {
    //freopen( "1.in" , "r" , stdin );
    //freopen( "1.out" , "w" , stdout );
    int t=1;
    //init();
    //s_1(t);
    for(int cas=1;cas<=t;cas++) {
        //printf("Case #%d: ",cas);
        solve();
    }
}