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]); }
Make Square CodeForces - 1028H (dp)
原文:https://www.cnblogs.com/uid001/p/11272123.html