bzoj 4028 : [HEOI2015]公约数数列

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<vector>
  6 #define N 100005
  7 #define d 200
  8 #define ll long long
  9 using namespace std;
 10 const int inf = 2000000000;
 11 int gcd(int a,int b)
 12 {
 13     if(!b)return a;
 14     return gcd(b,a%b);
 15 }
 16 int n;
 17 int a[N];
 18 int gd[N],xr[N],be[N];
 19 vector<int>g[N];
 20 vector<int>::iterator it;
 21 bool cmp(int x,int y)
 22 {
 23     if(xr[x]==xr[y])return x<y;
 24     return xr[x]<xr[y];
 25 }
 26 void gai(int x,int y)
 27 {
 28     int k=be[x];a[x]=y;
 29     int l=(k-1)*d+1,r=min(k*d,n);
 30     int xx=0,now=a[l];g[k].clear();
 31     for(int i=l;i<=r;i++)
 32     {
 33         now=gcd(now,a[i]);
 34         xx=xx^a[i];
 35         gd[i]=now;
 36         xr[i]=xx;
 37         g[k].push_back(i);
 38     }
 39     sort(g[k].begin(),g[k].end(),cmp);
 40 }
 41 void solve(ll x)
 42 {
 43     int now=a[1];int cnt=0,ed,xx=0;
 44     for(int i=1;i<=n;i+=d)
 45     {
 46         cnt++;ed=min(n,i+d-1);
 47         if(gd[ed]%now!=0)
 48         {
 49             for(int j=i;j<=ed;j++)
 50             {
 51                 if(gd[j]%now!=0)now=gcd(now,gd[j]);
 52                 if(x%now==0&&x/now==(ll)(xx^xr[j]))
 53                 {
 54                     printf("%d\n",j-1);
 55                     return ;
 56                 }
 57             }
 58         }
 59         else
 60         {
 61             if(x%now==0&&x/now<=inf)
 62             {
 63                 int tmp=x/now;
 64                 int l=0,r=g[cnt].size()-1;
 65                 tmp^=xx;
 66                 if(xr[g[cnt][r]]<tmp)
 67                 {
 68                     xx^=xr[ed];
 69                     continue;
 70                 }
 71                 while(l<r)
 72                 {
 73                     int mid=(l+r)>>1;
 74                     if(xr[g[cnt][mid]]>=tmp)r=mid;
 75                     else l=mid+1;
 76                 }
 77                 if(xr[g[cnt][r]]==tmp)
 78                 {
 79                     printf("%d\n",g[cnt][r]-1);
 80                     return ;
 81                 }
 82             }
 83         }
 84         xx^=xr[ed];
 85     }
 86     puts("no");
 87     return ;
 88 }
 89 int main()
 90 {
 91     scanf("%d",&n);
 92     for(int i=1;i<=n;i++)scanf("%d",&a[i]);
 93     int cnt=0;
 94     for(int i=1;i<=n;i+=d)
 95     {
 96         cnt++;
 97         int ed=min(n,i+d-1);
 98         int now=a[i];int xx=0;
 99         for(int j=i;j<=ed;j++)
100         {
101             be[j]=cnt;
102             now=gcd(now,a[j]);
103             xx=xx^a[j];
104             gd[j]=now;
105             xr[j]=xx;
106             g[cnt].push_back(j);
107         }
108         sort(g[cnt].begin(),g[cnt].end(),cmp);
109     }
110     int m;scanf("%d",&m);
111     char s[10];
112     int t1,t2;ll t3;
113     for(int i=1;i<=m;i++)
114     {
115         scanf("%s",s);
116         if(s[0]==‘M‘)
117         {
118             scanf("%d%d",&t1,&t2);
119             t1++;gai(t1,t2);
120         }
121         else
122         {
123             scanf("%lld",&t3);
124             solve(t3);
125         }
126     }
127     return 0;
128 }