bzoj 4026 dC Loves Number Theory

把我写吐了 太弱了
首先按照欧拉函数性质 我只需要统计区间不同质数个数就好了
一眼主席树
其次我被卡了分解质因数这里
可以通过质数筛时就建边解决
不够灵性啊,不知道如何改

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6+1000;
const int H = 5e4+5;
const int M = H*100;
const int mod = 1e6+777;

int isprime[N];
int prime[N]; int prnum[N]; int cc = 0;
int rev[N], has[N];
struct Node{
    int to,nex;
}E[N*3];
int head[N], eot = 0;
int n,q;
int a[H];
int T[H], ls[M], rs[M], v[M]; int tot;
int puf[H];

void add(int x,int y) {
    E[eot].to = y; E[eot].nex = head[x]; head[x] = eot++;
}
int build(int l,int r) {
    int rt = tot++;
    v[rt] = 1;
    if(l == r) return rt;
    int mid = (l+r)>>1;
    ls[rt] = build(l,mid); rs[rt] = build(mid+1,r);
    return rt;
}
int upd(int pos,int num,int l,int r,int pr) {
    int rt = tot++;
    v[rt] = 1ll*v[pr]*num%mod;
    if(l == r) return rt;
    int mid = (l+r)>>1;
    if(pos <= mid) {
        ls[rt] = upd(pos,num,l,mid,ls[pr]); rs[rt] = rs[pr];
    }else {
        ls[rt] = ls[pr]; rs[rt] = upd(pos,num,mid+1,r,rs[pr]);
    }
    return rt;
}
int query(int L,int R,int l,int r,int rt) {
    if(L <= l && r <= R) return v[rt];
    int mid = (l+r)>>1;
    int ans = 1;
    if(L <= mid) ans = 1ll*ans*query(L,R,l,mid,ls[rt]) %mod;
    if(R > mid)  ans = 1ll*ans*query(L,R,mid+1,r,rs[rt]) %mod;
    return ans;
}
void debug(int l,int r,int rt) {
    printf("%d %d %d
",l,r,v[rt]);
    if(l == r) return;
    int mid  = (l+r)>>1;
    debug(l,mid,ls[rt]); debug(mid+1, r, rs[rt]);
}


int main(){
    memset(head,-1,sizeof(head));
    for(int i = 2; i < N; ++i) {
        if(!isprime[i]) {
            prime[++cc] = i;
            for(int j = i; j < N; j += i) {
                isprime[j] = 1;
                add(j,i); 
            }
        }
    }
    rev[0] = rev[1] = 1;
    for(int i=2;i< N;++i) {
        rev[i]= 1ll*(mod-mod/i)* rev[mod%i]%mod;
    }
    for(int i = 1; i <= cc; ++i) {
        prnum[prime[i]] = 1ll*(prime[i]-1) * rev[prime[i]] %mod;
    }

    while(~scanf("%d %d",&n,&q)) {
        memset(has,0,sizeof(has));
        for(int i = 1; i <= n; ++i) scanf("%d",&a[i]);
        tot = 0; puf[0] = 1;
        T[0] = build(1,n);  
        for(int i = 1; i <= n; ++i) {
            puf[i] = 1ll*puf[i-1]*a[i]%mod;
            int tmp = T[i-1];
            int tt = 1;
            for(int j = head[a[i]]; j != -1; j = E[j].nex) {
                int x = E[j].to;
                if(has[x]) 
                    tmp = upd(has[x],rev[prnum[x]],1,n,tmp);
                tt = 1ll*tt*prnum[x]%mod; 
                has[x] = i;
            }   
            T[i] = upd(i,tt,1,n,tmp);
        }

        int ans = 0;
        for(int i = 0; i < q; ++i) {
            int l,r; scanf("%d %d",&l,&r);
            l^=ans; r^=ans;
            ans = query(l,r,1,n,T[r]);
            ans = 1ll*ans*puf[r]%mod *rev[puf[l-1]] %mod;
            printf("%d
",ans);
        }

    }
    return 0;   
}