zcmimi's blog
查看原题

点击跳转

我们可以发现gcd(a_l,...,a_r)是单调递减的,而or(a_l,...,a_r)是单调递增的

所以我们可以两次二分来找到合法区间

至于求gcd(a_l,...,a_r),or(a_l,...,a_r)我们可以用rmq来求

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;rg char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;rg char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;rg char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;rg char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
#define N 500011
int n,k,a[N],f[21][N],g[21][N];
void st(){
    Fur(i,1,n)f[0][i]=g[0][i]=a[i];
    Fur(i,1,20)
        Fur(j,1,n-(1<<i)+1){
            f[i][j]=f[i-1][j]|f[i-1][j+(1<<(i-1))];
            g[i][j]=GCD(g[i-1][j],g[i-1][j+(1<<(i-1))]);
        }
}
int g1(int l,int r){
    int sz=l2(r-l+1);
    return f[sz][l]|f[sz][r-(1<<sz)+1];
}
int g2(int l,int r){
    int sz=l2(r-l+1);
    return GCD(g[sz][l],g[sz][r-(1<<sz)+1]);
}
ll ans=0;
int find(int s,int x,int l,int r){
    int as;
    while(l<=r){
        int m=(l+r)>>1,z=g2(s,m);
        if(x==z)as=m,l=m+1;
        else r=m-1;
    }
    return as;
}
void solve(int s,int x,int l,int r){
    int L=l,R=r,num=0,Num=-1; 
    while(l<=r){
        int m=(l+r)>>1,z=g1(s,m);
        if(x==z)r=m-1,num=m; 
        else if(x<z)r=m-1;
        else if(x>z)l=m+1;
    }
    while(L<=R){
        int m=(L+R)>>1,z=g1(s,m);
        if(x==z)L=m+1,Num=m;
        else if(x<z)R=m-1;
        else if(x>z)L=m+1;
    }
    ans+=Num-num+1;
}
int main(){
    in(n,k);
    Fur(i,1,n)in(a[i]);
    st();
    Fur(i,1,n)
        Fur(j,i,n){
            int t=g2(i,j),p=find(i,t,j,n);
            solve(i,t^k,j,p);
            j=p;
        }
    printf("%lld\n",ans);
}
LG 3794 签到题IV
comment评论
Search
search