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; }