New Year Concert CodeForces - 1632D

http://codeforces.com/contest/1632/problem/D

"d can be any positive integer" is the key: since any number can be selected, a prime number greater than 1e9 can be selected.

The line segment tree is used for interval query, and the AC code complexity is n*logn*logn.

Among them, logn*logn is a bisection*interval query. You can omit the bisection by modifying the interval query (the line segment tree is constantly bisection), reducing the complexity to n*logn.

#include <bits/stdc++.h>
using namespace std;
const int maxn=2e5+10;

int tree[4*maxn];
int ary[maxn];
int n;

void pushup(int cur)
{
    tree[cur]=__gcd(tree[2*cur],tree[2*cur+1]);
}

void build(int l,int r,int cur)
{
    int mid;
    if(l==r){
        tree[cur]=ary[l];
        return;
    }
    mid=(l+r)/2;
    build(l,mid,2*cur);
    build(mid+1,r,2*cur+1);
    pushup(cur);
}

int query(int pl,int pr,int l,int r,int cur)
{
    int mid,resl,resr;
    if(pl<=l&&r<=pr){
        return tree[cur];
    }
    mid=(l+r)/2,resl=-1,resr=-1;
    if(pl<=mid) resl=query(pl,pr,l,mid,2*cur);
    if(pr>mid) resr=query(pl,pr,mid+1,r,2*cur+1);
    if(resl==-1) return resr;
    else if(resr==-1) return resl;
    else return __gcd(resl,resr);
}

int main()
{
    int i,p,res,ans;
    int l,r,mid,tar;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d",&ary[i]);
    }
    build(1,n,1);
    p=0,ans=0;
    for(i=1;i<=n;i++){
        l=p+1,r=i,tar=-1;
        while(l<=r){
            mid=(l+r)/2;
            res=query(mid,i,1,n,1);
            if(res>i-mid+1){
                r=mid-1;
            }
            else if(res<i-mid+1){
                l=mid+1;
            }
            else{
                tar=mid;
                break;
            }
        }
        if(tar!=-1){
            p=i,ans++;
        }
        printf("%d ",ans);
    }
    printf("\n");
    return 0;
}