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<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),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
";
#else
    const char LL[]="%lld
";
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i]=read();
    prepos[0]=0;
    for (int i=1;i<=n;i++)
    if (gcd(a[i],prenum[precnt])!=prenum[precnt]) precnt++,prepos[precnt]=i,prenum[precnt]=gcd(a[i],prenum[precnt-1]);
    prepos[precnt+1]=n+1;
    sufpos[0]=n+1;
    for (int i=n;i>=1;i--)
    if (gcd(a[i],sufnum[sufcnt])!=sufnum[sufcnt]) sufcnt++,sufpos[sufcnt]=i,sufnum[sufcnt]=gcd(a[i],sufnum[sufcnt-1]);
    sufpos[sufcnt+1]=0;
    for (int i=0;i<=precnt;i++)
        for (int j=0;j<=sufcnt;j++)
        {
            int x=gcd(prenum[i],sufnum[j]);
            for (int k=prepos[i];k<prepos[i+1];k++) 
            ans=(ans+1ll*x*max(0,sufpos[j]-max(sufpos[j+1],k+1))%P)%P;
        }
    ans-=a[1],ans-=a[n];while (ans<0) ans+=P;
    cout<<ans;
    return 0;
}