BZOJ4921 互质序列

　　即求删掉一个子序列的gcd之和。注意到前后缀gcd的变化次数都是log级的，于是暴力枚举前缀gcd和后缀gcd即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
#define P 998244353
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,a[N],prepos[N],prenum[N],precnt,sufpos[N],sufnum[N],sufcnt,ans;
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4921.in","r",stdin);
    freopen("bzoj4921.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=;i<=n;i++) a[i]=read();
    prepos[]=;
    for (int i=;i<=n;i++)
    if (gcd(a[i],prenum[precnt])!=prenum[precnt]) precnt++,prepos[precnt]=i,prenum[precnt]=gcd(a[i],prenum[precnt-]);
    prepos[precnt+]=n+;
    sufpos[]=n+;
    for (int i=n;i>=;i--)
    if (gcd(a[i],sufnum[sufcnt])!=sufnum[sufcnt]) sufcnt++,sufpos[sufcnt]=i,sufnum[sufcnt]=gcd(a[i],sufnum[sufcnt-]);
    sufpos[sufcnt+]=;
    for (int i=;i<=precnt;i++)
        for (int j=;j<=sufcnt;j++)
        {
            int x=gcd(prenum[i],sufnum[j]);
            for (int k=prepos[i];k<prepos[i+];k++)
            ans=(ans+1ll*x*max(,sufpos[j]-max(sufpos[j+],k+))%P)%P;
        }
    ans-=a[],ans-=a[n];while (ans<) ans+=P;
    cout<<ans;
    return ;
}