Make Square CodeForces - 1028H (dp)

大意: 若一个序列存在两个数的积为完全平方数, 则为好序列. 给定序列$a$, 每次询问求子区间$[l,r]$最少修改多少次可以成为好序列, 每次修改可以任选素数$p$, 任选一个数乘或除$p$.

$dp_{x,y}$表示状态为$x$删除$y$个因子的最大位置

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

const int N = 1e7+10;
int n, q, ans[N];
int ss[20], dp[N][10], L[N];
vector<int> g[N], p[N];

int main() {
	scanf("%d%d", &n, &q);
	REP(i,1,n) {
		int x;
		scanf("%d", &x);
		int mx = sqrt(x+0.5);
		REP(j,2,mx) if (x%j==0) {
			int t = 0;
			while (x%j==0) t^=1,x/=j;
			if (t) p[i].pb(j);
		}
		if (x) p[i].pb(x);
	}
	REP(i,1,q) {
		int r;
		scanf("%d%d",L+i,&r);
		g[r].pb(i);
	}
	REP(i,1,n) {
		int sz = p[i].size(), mx = (1<<sz)-1;
		REP(s,0,mx) {
			int x = 1, y = 0;
			REP(j,0,sz-1) {
				if (s>>j&1) x*=p[i][j];
				else ++y;
			}
			REP(j,0,15) ss[j+y]=max(ss[j+y],dp[x][j]);
		}
		for (auto j:g[i]) {
			int now = 0;
			while (ss[now]<L[j]) ++now;
			ans[j] = now;
		}
		REP(s,0,mx) {
			int x = 1, y = 0;
			REP(j,0,sz-1) {
				if (s>>j&1) x*=p[i][j];
				else ++y;
			}
			dp[x][y] = i;
		}
	}
	REP(i,1,q) printf("%d\n",ans[i]);
}