[jzoj4711]【NOIP2016提高A组模拟8.17】Binary

Description

Solution

Code

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define fo(i,x,y) for (ll pq = (y),i = (x);i <= pq;++ i)
#define fd(i,x,y) for (ll pq = (y),i = (x);i >= pq;-- i)
#define oo 2139062143
using namespace std;
typedef double db;
typedef long long ll;
ll lowbit(ll x) {return((x)&(-x));}
ll min(ll x,ll y){return (x>y)?(y):(x);}
ll max(ll x,ll y){return (x>y)?(x):(y);}
ll er(ll x)
{
return(1<<(x));
}
const ll N=100100,MX=(1<<21)-1;
ll n,q;
ll a
;
ll opt,x,y;
ll t[22][MX];
void add(ll pos,ll x,ll v)
{
++x;
while(x<=MX)
{
t[pos][x]+=v;
x+=lowbit(x);
}
}
ll query(ll pos,ll x)
{
++x;
x=min(x,MX);
ll rt=0;
while(x>0)
{
rt+=t[pos][x];
x-=lowbit(x);
}
return rt;
}
int main()
{
scanf("%lld%d",&n,&q);
fo(i,1,n)
{
scanf("%lld",&a[i]);
fo(l,0,20) add(l,a[i]&(er(l+1)-1),1);
}
while(q--)
{
scanf("%lld%lld%lld",&opt,&x,&y);
if(opt==1)
{
fo(l,0,20) add(l,a[x]&(er(l+1)-1),-1),add(l,y&(er(l+1)-1),1);
a[x]=y;
}
else
if(opt==2)
{
ll ans=0;
fo(l,0,20)
if(y&er(l))
{
ll tmp=0;
tmp+=(query(l,er(l+1)-1-(x&(er(l+1)-1)))-query(l,er(l)-1-(x&(er(l+1)-1))));
tmp+=query(l,er(l+1)+er(l+1)-1-(x&(er(l+1)-1)))-query(l,er(l+1)+er(l)-1-(x&(er(l+1)-1)));
ans+=tmp*er(l);
}
printf("%lld\n",ans);
}
}
return 0;
}