BZOJ4028 HEOI2015公约数数列（分块）

　　前缀gcd的变化次数是log的，考虑对每一种gcd查询，问题变为查询一段区间是否存在异或前缀和=x/gcd。

　　无修改的话显然可以可持久化trie，但这玩意实在没法支持修改。于是考虑分块。

　　对于每一块将其中所有块内异或前缀和排序。查询时先看这块与上一块相比gcd有没有变化，如果有对其中每个位置暴力查询，否则在排序后的数组中二分。修改时暴力改每一块的前缀gcd及异或和，被修改的块暴力重构排序数组即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100010
#define ll long long
ll read()
{
    ll 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,m,a[N];
int block_size,block_tot,L[N],R[N],pos[N];
int gcd_pre[N],xor_pre[N],gcd_block[N],xor_block[N];
struct data
{
    int x,i;
    bool operator <(const data&a) const
    {
        return x<a.x||x==a.x&&i<a.i;
    }
}v[N];
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4028.in","r",stdin);
    freopen("bzoj4028.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i]=read();
    block_size=sqrt(n);block_tot=(n-1)/block_size+1;
    for (int i=1;i<=block_tot;i++)
    L[i]=(i-1)*block_size+1,R[i]=min(n,i*block_size);
    for (int i=1;i<=block_tot;i++)
    {
        for (int j=L[i];j<=R[i];j++)
        gcd_block[i]=gcd(gcd_block[i],a[j]),
        xor_block[i]=xor_block[i]^a[j],
        pos[j]=i;
        gcd_pre[i]=gcd(gcd_pre[i-1],gcd_block[i]),xor_pre[i]=xor_pre[i-1]^xor_block[i];
    }
    for (int i=1;i<=block_tot;i++)
    {
        v[L[i]].x=a[L[i]],v[L[i]].i=L[i];
        for (int j=L[i]+1;j<=R[i];j++)
        v[j].x=v[j-1].x^a[j],v[j].i=j;
        sort(v+L[i],v+R[i]+1);
    }
    m=read();
    while (m--)
    {
        char c=getchar();
        while (c<'A'||c>'Z') c=getchar();
        if (c=='M')
        {
            int p=read()+1,x=read();a[p]=x;p=pos[p];
            gcd_block[p]=xor_block[p]=0;
            for (int i=L[p];i<=R[p];i++)
            gcd_block[p]=gcd(gcd_block[p],a[i]),
            xor_block[p]=xor_block[p]^a[i];
            for (int i=p;i<=block_tot;i++)
            gcd_pre[i]=gcd(gcd_pre[i-1],gcd_block[i]),
            xor_pre[i]=xor_pre[i-1]^xor_block[i];
            v[L[p]].x=a[L[p]],v[L[p]].i=L[p];
            for (int i=L[p]+1;i<=R[p];i++)
            v[i].x=v[i-1].x^a[i],v[i].i=i;
            sort(v+L[p],v+R[p]+1);
        }
        else
        {
            ll x=read();int ans=0;
            for (int i=1;i<=block_tot;i++)
            {
                if (ans) break;
                if (gcd_pre[i]==gcd_pre[i-1])
                {
                    if (x%gcd_pre[i-1]==0)
                    {
                        int l=L[i],r=R[i];
                        while (l<=r)
                        {
                            int mid=l+r>>1;
                            if (v[mid].x>=(x/gcd_pre[i-1]^xor_pre[i-1])) ans=mid,r=mid-1;
                            else l=mid+1;
                        }
                        if (ans&&v[ans].x==(x/gcd_pre[i-1]^xor_pre[i-1])) {ans=v[ans].i;break;}
                        else ans=0;
                    }
                }
                else
                {
                    int p=gcd_pre[i-1],q=xor_pre[i-1];
                    for (int j=L[i];j<=R[i];j++)
                    {
                        p=gcd(p,a[j]),q^=a[j];
                        if (x%p==0&&x/p==q) {ans=j;break;}
                    }
                }
            }
            if (ans) printf("%d\n",ans-1);
            else printf("no\n");
        }
    }
    return 0;
}