This problem solution is willing to Los scum of fresh small valley P3674

Mo + bitset version of the sub-title team

Too much nonsense questions surface, causing discomfort.

For one, two and asked, with a bitset maintain each number in the range appears, and the feasibility of judgment or misplaced.

The third number to ask about before simply on the basis of enumeration (complexity root n)

The total complexity of O (n ^ 1.5)

(My code is ugly giant, think no need to write separate, I fear a fool)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<bitset>
using namespace std;

const int maxn = 200005;
int n, m, num[maxn];
int ans[maxn], n1, n2, n3;

struct ss {
    int l;
    int r;
    int x;
    int p;
} a[maxn], a1[maxn], a2[maxn], a3[maxn];

bool cmp_r(ss a, ss b) {
    return a.r < b.r;
}

bool cmp_l(ss a, ss b) {
    return a.l < b.l;
}

int nl[maxn], nr[maxn], tnr[maxn], ss[maxn];
bitset <maxn> s;
bitset <maxn> s2;
bitset <maxn> st;

int count1() {
    memset(nl, 0, sizeof(nl));
    memset(nr, 0, sizeof(nr));
    memset(tnr, 0, sizeof(tnr));
    int i, j, q, ll, rr;
    int tn=0, sizek;
    sort(a+1, a+1+m, cmp_l);
    sizek=sqrt(n);
    for(i=1; i <= m; i++) 
        tnr[a[i].l/sizek+1]=i;
    for(i=1; i <= n; i++) {
        if(tnr[i] != 0) tn++, nl[tn]=nr[tn-1]+1, nr[tn]=tnr[i];
    }
    
    for(i=1; i <= tn; i++) {
        memset(ss, 0, sizeof(ss));
        s.reset();
        st.reset();
        sort(a+nl[i], a+nr[i]+1, cmp_r);
        ll=a[nl[i]].l; rr=a[nl[i]].l-1;
        
        for(q=nl[i]; q <= nr[i]; q++)  {
            for(j=rr+1; j <= a[q].r; j++) {
                s[num[j]]=1;
                ss[num[j]]++;
            }
            for(j=a[q].l; j <= ll-1; j++) {
                s[num[j]]=1;
                ss[num[j]]++;
            }
            for(j=ll; j <= a[q].l-1; j++) {
                ss[num[j]]--;
                if(!ss[num[j]]) s[num[j]]=0;
            }
            ll=a[q].l; rr=a[q].r;
            st=((s<<a[q].x) & (s));
            if(st.any()) ans[a[q].p]=1;
            else ans[a[q].p]=0;
        }
    }
    return 0;
}

int count2() {
    memset(nl, 0, sizeof(nl));
    memset(nr, 0, sizeof(nr));
    memset(tnr, 0, sizeof(tnr));
    int i, j, q, ta, ll, rr;
    int tn=0, sizek;
    sort(a+1, a+1+m, cmp_l);
    sizek=sqrt(n);
    for(i=1; i <= m; i++) 
        tnr[a[i].l/sizek+1]=i;
    for(i=1; i <= n; i++) {
        if(tnr[i] != 0) tn++, nl[tn]=nr[tn-1]+1, nr[tn]=tnr[i];
    }
    
    for(i=1; i <= tn; i++) {
        memset(ss, 0, sizeof(ss));
        s.reset();
        st.reset();
        s2.reset();
        sort(a+nl[i], a+nr[i]+1, cmp_r);
        ll=a[nl[i]].l; rr=a[nl[i]].l-1;
        
        for(q=nl[i]; q <= nr[i]; q++)  {
            for(j=rr+1; j <= a[q].r; j++) {
                s[num[j]]=1;
                s2[100000-num[j]]=1;
                ss[num[j]]++;
            }
            for(j=a[q].l; j <= ll-1; j++) {
                s[num[j]]=1;
                s2[100000-num[j]]=1;
                ss[num[j]]++;
            }
            for(j=ll; j <= a[q].l-1; j++) {
                ss[num[j]]--;
                if(!ss[num[j]]) s[num[j]]=0, s2[100000-num[j]]=0;
            }
            ll=a[q].l; rr=a[q].r;
            st=((s<<(100000-a[q].x)) & (s2));
            if(st.any()) ans[a[q].p]=1;
            else ans[a[q].p]=0;
        }
    }
    return 0;
}

int count3() {
    memset(nl, 0, sizeof(nl));
    memset(nr, 0, sizeof(nr));
    memset(tnr, 0, sizeof(tnr));
    int i, j, q, ta, ll, rr, y;
    int tn=0, sizek;
    sort(a+1, a+1+m, cmp_l);
    sizek=sqrt(n);
    for(i=1; i <= m; i++) 
        tnr[a[i].l/sizek+1]=i;
    for(i=1; i <= n; i++) {
        if(tnr[i] != 0) tn++, nl[tn]=nr[tn-1]+1, nr[tn]=tnr[i];
    }
    
    for(i=1; i <= tn; i++) {
        memset(ss, 0, sizeof(ss));
        s.reset();
        sort(a+nl[i], a+nr[i]+1, cmp_r);
        ll=a[nl[i]].l; rr=a[nl[i]].l-1;
        
        for(q=nl[i]; q <= nr[i]; q++)  {
            for(j=rr+1; j <= a[q].r; j++) {
                s[num[j]]=1;
                ss[num[j]]++;
            }
            for(j=a[q].l; j <= ll-1; j++) {
                s[num[j]]=1;
                ss[num[j]]++;
            }
            for(j=ll; j <= a[q].l-1; j++) {
                ss[num[j]]--;
                if(!ss[num[j]]) s[num[j]]=0;
            }
            ll=a[q].l; rr=a[q].r;
            
            ans[a[q].p]=0;
            for(y=1; y <= sqrt(a[q].x); y++) {
                if(a[q].x%y == 0 && s[y] && s[a[q].x/y]) ans[a[q].p]=1;
            }
        }
    }
    return 0;
}


int main() {
    ios::sync_with_stdio(false);
    int i, tm, ta;
    cin>>n>>m;
    for(i=1; i <= n; i++) 
        cin>>num[i];
    for(i=1; i <= m; i++) {
        cin>>ta;
        if(ta == 1) n1++, cin>>a1[n1].l>>a1[n1].r>>a1[n1].x, a1[n1].p=i;
        if(ta == 2) n2++, cin>>a2[n2].l>>a2[n2].r>>a2[n2].x, a2[n2].p=i;
        if(ta == 3) n3++, cin>>a3[n3].l>>a3[n3].r>>a3[n3].x, a3[n3].p=i;
    }
    
    tm=m;
    m=n1; for(i=1; i <= m; i++) a[i]=a1[i]; count1();
    m=n2; for(i=1; i <= m; i++) a[i]=a2[i]; count2();
    m=n3; for(i=1; i <= m; i++) a[i]=a3[i]; count3();
    
    for(i=1; i <= tm; i++) if(ans[i]) cout<<"hana"<<endl; else cout<<"bi"<<endl;
    return 0;
}