CF359D Pair of Numbers gcd+暴力

利用区间 gcd 个数不超过 log 种来做就可以了~

code:

#include <bits/stdc++.h>
#define N 300005
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
int n;
set<int>s;
struct solve
{
	int A[N],pos[N],len[N],gc[N],b[N],top,tmp;
	void get()
	{
		int i,j;
		for(i=1;i<=n;++i)
		{
			for(j=1;j<=top;++j) gc[j]=__gcd(gc[j], A[i]);
			gc[++top]=A[i], pos[top]=i, b[tmp=1]=1;
			for(j=2;j<=top;++j) if(gc[j]!=gc[j-1]) b[++tmp]=j;
			for(top=0,j=1;j<=tmp;++j) gc[++top]=gc[b[j]], pos[top]=pos[b[j]];
			len[i]=i-pos[top]+1;
		}
	}
}ori,rev;
int main()
{
	// setIO("input");
	int i,j,re=0,cnt=0;
	scanf("%d",&n);
	for(i=1;i<=n;++i) scanf("%d",&ori.A[i]);
	for(i=1;i<=n;++i) rev.A[i]=ori.A[n-i+1];
	ori.get(), rev.get();
	for(i=1;i<=n;++i) re=max(re, ori.len[i]+rev.len[n-i+1]-2);
	for(i=1;i<=n;++i) if(ori.len[i]+rev.len[n-i+1]-2==re) s.insert(i-ori.len[i]+1);
	printf("%d %d\n",s.size(), re);
	for(set<int>::iterator it=s.begin();it!=s.end();it++) printf("%d ",*it);
	return 0;
}