P3709 uncle string title (Mo + team concluded)

topic

P3709 uncle's title string

practice

There is a obvious conclusion: the smallest for a range of answers to the mode number

Mo with the team to find the number of the congregation offline

\ (tmp_i \) indicates that an \ (i \) the number of number of times, \ (num_i \) represents \ (i \) number of occurrences

Narrow range: the answer may be reduced, to see where the answer \ (tmp \) if not unique

Expand the range: The answer is increased

Code

#include<bits/stdc++.h>
typedef int LL;
const LL maxn=1e6+9;
inline LL Read(){
    LL x(0),f(1); char c=getchar();
    while(c<'0' || c>'9'){
        if(c=='-') f=-1; c=getchar();
    }
    while(c>='0' && c<='9'){
        x=(x<<3ll)+(x<<1ll)+c-'0'; c=getchar();
    }return x*f;
}
struct node{
    LL l,r,id;
}qy[maxn];
LL n,m,ret;
LL ans[maxn],num[maxn],tmp[maxn],a[maxn],b[maxn],bel[maxn];
inline bool cmp(node xx,node yy){
    return bel[xx.l]<bel[yy.l] || (bel[xx.l]==bel[yy.l] && (bel[xx.l]&1?xx.r<yy.r:xx.r>yy.r));
}
inline void Modify(LL val,LL op){
    if(!op){
        if(ret==num[val] && tmp[ret]==1) --ret;
        --tmp[num[val]]; ++tmp[--num[val]];
    }else{
        if(ret==num[val]) ++ret;
        --tmp[num[val]]; ++tmp[++num[val]];
    }
}
int main(){
    n=Read(); m=Read();
    for(LL i=1;i<=n;++i) a[i]=b[i]=Read();
    std::sort(b+1,b+1+n);
    for(LL i=1;i<=n;++i) a[i]=std::lower_bound(b+1,b+1+n,a[i])-b;//,printf("%d ",a[i]);puts("");
    for(LL i=1;i<=m;++i) qy[i]=(node){Read(),Read(),i};
    LL pieces(sqrt(n)),size(n/pieces);
    for(LL i=1;i<=pieces;++i){
        for(LL j=(i-1)*size+1;j<=i*size;++j)
            bel[j]=i;
    }
    for(LL i=pieces*size+1;i<=n;++i)
        bel[i]=pieces+1;
    std::sort(qy+1,qy+1+m,cmp);
    tmp[0]=n;
    LL nl(1),nr(0);
    for(LL i=1;i<=m;++i){
        LL ql(qy[i].l),qr(qy[i].r);
        while(nl<ql) Modify(a[nl++],0);
        while(nl>ql) Modify(a[--nl],1);
        while(nr<qr) Modify(a[++nr],1);
        while(nr>qr) Modify(a[nr--],0);
        ans[qy[i].id]=ret;
    }
    for(LL i=1;i<=m;++i) printf("-%d\n",ans[i]);
    return 0;
}