题解:【CF1513D】 GCD and MST
更好理解的无脑方法。要寻找区间使得 \(\gcd_{i = l}^{r} a_i = \min_{i = l}^{r} a_i\),这意味着区间内所有值都为最小值的倍数。假设最小值为 \(a_k (l \leq k \leq r)\),那么我们把最小值放在边上,这样就可以拆成正序找一遍再倒序找一遍,即 \(\forall a_i (l \leq i \leq k),a_k \mid a_i\) 和 \(\forall a_i (k \leq i \leq r),a_k \mid a_i\)。初始 \(a_i\) 和 \(a_{i+1}\) 都相连权为 \(p\) 的边,找出所有区间,每个区间意味着区间内的边都可以替换成找到的最小值,将区间按最小值从大到小排序后按顺序覆盖即可,因为后覆盖的一定比先覆盖的优。维护区间和,区间推平,暴力线段树即可。因为是最小生成树,所以代码实现上 \(i\) 上的值是 \(i\) 和 \(i+1\) 间的连边边权,于是做到 \(\mathcal O(n \log n)\)。
#include<bits/stdc++.h>
#define ld long double
#define ui unsigned int
#define ull unsigned long long
#define int long long
#define eb emplace_back
#define pb pop_back
#define ins insert
#define mp make_pair
#define pii pair<int,int>
#define fi first
#define se second
#define power(x) ((x)*(x))
using namespace std;
namespace FastIO
{
template<typename T=int> inline T read()
{
T s=0,w=1; char c=getchar();
while(!isdigit(c)) {if(c=='-') w=-1; c=getchar();}
while(isdigit(c)) s=(s*10)+(c^48),c=getchar();
return s*w;
}
template<typename T> inline void read(T &s)
{
s=0; int w=1; char c=getchar();
while(!isdigit(c)) {if(c=='-') w=-1; c=getchar();}
while(isdigit(c)) s=(s*10)+(c^48),c=getchar();
s=s*w;
}
template<typename T,typename... Args> inline void read(T &x,Args &...args)
{
read(x),read(args...);
}
template<typename T> inline void write(T x,char ch)
{
if(x<0) x=-x,putchar('-');
static char stk[25]; int top=0;
do {stk[top++]=x%10+'0',x/=10;} while(x);
while(top) putchar(stk[--top]);
if(ch!='~') putchar(ch);
return;
}
}
using namespace FastIO;
namespace MTool
{
#define TA template<typename T,typename... Args>
#define TT template<typename T>
static const int Mod=998244353;
TT inline void Swp(T &a,T &b) {T t=a;a=b;b=t;}
TT inline void cmax(T &a,T b) {a=max(a,b);}
TT inline void cmin(T &a,T b) {a=min(a,b);}
TA inline void cmax(T &a,T b,Args... args) {a=max({a,b,args...});}
TA inline void cmin(T &a,T b,Args... args) {a=min({a,b,args...});}
TT inline void Madd(T &a,T b) {a=a+b>=Mod?a+b-Mod:a+b;}
TT inline void Mdel(T &a,T b) {a=a-b<0?a-b+Mod:a-b;}
TT inline void Mmul(T &a,T b) {a=a*b%Mod;}
TT inline void Mmod(T &a) {a=(a%Mod+Mod)%Mod;}
TT inline T Cadd(T a,T b) {return a+b>=Mod?a+b-Mod:a+b;}
TT inline T Cdel(T a,T b) {return a-b<0?a-b+Mod:a-b;}
TT inline T Cmul(T a,T b) {return a*b%Mod;}
TT inline T Cmod(T a) {return (a%Mod+Mod)%Mod;}
TA inline void Madd(T &a,T b,Args... args) {Madd(a,Cadd(b,args...));}
TA inline void Mdel(T &a,T b,Args... args) {Mdel(a,Cadd(b,args...));}
TA inline void Mmul(T &a,T b,Args... args) {Mmul(a,Cmul(b,args...));}
TA inline T Cadd(T a,T b,Args... args) {return Cadd(Cadd(a,b),args...);}
TA inline T Cdel(T a,T b,Args... args) {return Cdel(Cdel(a,b),args...);}
TA inline T Cmul(T a,T b,Args... args) {return Cmul(Cmul(a,b),args...);}
TT inline T qpow(T a,T b) {int res=1; while(b) {if(b&1) Mmul(res,a); Mmul(a,a); b>>=1;} return res;}
TT inline T qmul(T a,T b) {int res=0; while(b) {if(b&1) Madd(res,a); Madd(a,a); b>>=1;} return res;}
TT inline T spow(T a,T b) {int res=1; while(b) {if(b&1) res=qmul(res,a); a=qmul(a,a); b>>=1;} return res;}
TT inline void exgcd(T A,T B,T &X,T &Y) {if(!B) return X=1,Y=0,void(); exgcd(B,A%B,Y,X),Y-=X*(A/B);}
TT inline T Ginv(T x) {T A=0,B=0; exgcd(x,Mod,A,B); return Cmod(A);}
#undef TT
#undef TA
}
using namespace MTool;
inline void file()
{
freopen(".in","r",stdin);
freopen(".out","w",stdout);
return;
}
bool Mbe;
namespace LgxTpre
{
static const int MAX=200010;
static const int inf=2147483647;
static const int INF=4557430888798830399;
int T,n,p,a[MAX];
struct lmy{int l,r,v;}b[MAX]; int cnt;
namespace SegmentTree
{
int tag[MAX<<2],sum[MAX<<2];
inline void pushup(int i) {sum[i]=sum[i<<1]+sum[i<<1|1];}
inline void down(int i,int l,int r,int k) {sum[i]=(r-l+1)*k,tag[i]=k;}
inline void pushdown(int i,int l,int r)
{
if(!tag[i]||l==r) return;
int mid=(l+r)>>1;
down(i<<1,l,mid,tag[i]),down(i<<1|1,mid+1,r,tag[i]),tag[i]=0;
}
void build(int i,int l,int r)
{
tag[i]=0;
if(l==r) return sum[i]=p,void();
int mid=(l+r)>>1;
build(i<<1,l,mid),build(i<<1|1,mid+1,r);
pushup(i);
}
void modify(int i,int l,int r,int L,int R,int k)
{
if(l<=L&&R<=r) return down(i,L,R,k);
pushdown(i,L,R);
int mid=(L+R)>>1;
if(l<=mid) modify(i<<1,l,r,L,mid,k);
if(r>mid) modify(i<<1|1,l,r,mid+1,R,k);
pushup(i);
}
}
using namespace SegmentTree;
inline void lmy_forever()
{
read(T);
while(T--)
{
read(n,p),cnt=0,build(1,1,n-1);
for(int i=1;i<=n;++i) read(a[i]);
for(int l=1,r=1;r<=n;l=r+1,++r) {if(a[l]>=p) continue; while(r<=n&&a[r]%a[l]==0) ++r; --r; if(l!=r) b[++cnt]=(lmy){l,r-1,a[l]};}
for(int l=n,r=n;l;r=l-1,--l) {if(a[r]>=p) continue; while(l>=1&&a[l]%a[r]==0) --l; ++l; if(l!=r) b[++cnt]=(lmy){l,r-1,a[r]};}
sort(b+1,b+cnt+1,[](lmy x,lmy y){return x.v>y.v;});
for(int i=1;i<=cnt;++i) modify(1,b[i].l,b[i].r,1,n-1,b[i].v);
write(sum[1],'\n');
}
}
}
bool Med;
signed main()
{
// file();
fprintf(stderr,"%.3lf MB\n",abs(&Med-&Mbe)/1048576.0);
int Tbe=clock();
LgxTpre::lmy_forever();
int Ted=clock();
cerr<<1e3*(Ted-Tbe)/CLOCKS_PER_SEC<<" ms\n";
return (0-0);
}