nowcoder15731 I. Five Day Couple

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题解

可持久化Trie树裸题

把原先的数转化成二进制然后建立可持久化Trie

每次查询，贪心在Trie树上走动即可

代码

#include 
#include 
#include 
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct PersistenceTrie
{
    #define SIZE 2
    int tot, ch[maxn<<5][SIZE], end[maxn<<5], root[maxn], rtot, cnt[maxn<<5];
    void init()
    {
        tot=1;
        cl(ch[tot]), cnt[tot]=1;
        root[rtot=0]=1;
    }
    int New()
    {
        tot++;
        cl(ch[tot]);
        end[tot]=0;
        cnt[tot]=1;
        return tot;
    }
    int Copy(int x)
    {
        tot++;
        memcpy(ch[tot],ch[x],sizeof(ch[x]));
        end[tot]=end[x];
        cnt[tot]=cnt[x]+1;
        return tot;
    }
    int* operator[](int index){return ch[index];}
    int insert(int* s, int len)
    {
        root[rtot+1] = Copy(root[rtot]);
        rtot++;
        int u=root[rtot], i;
        for(i=1;i<=len;i++)
        {
            int &v = ch[u][s[i]];
            u=v= v?Copy(v):New();
        }
        end[u]++;
        return u;
    }
}T;
int main()
{
    ll n=read(), m, i, j;
    T.init();
    rep(i,1,n)
    {
        ll a=read();
        int x[50];
        rep(j,0,31)
        {
            x[32-j] = !!( (1ll<<j) & a );
        }
        T.insert(x,32);
    }
    de(T.rtot);
    m=read();
    while(m--)
    {
        ll b=read(), l=read(), r=read();
        ll ul=T.root[l-1], ur=T.root[r];
        drep(i,31,0)
        {
            ll d = !!( (1ll<<i) & b );
            if(T.cnt[T[ur][!d]] - T.cnt[T[ul][!d]])
            {
                ul=T[ul][!d];
                ur=T[ur][!d];
                b |= 1ll<<i;
            }
            else
            {
                ul=T[ul][d];
                ur=T[ur][d];
                b &= ~(1ll<<i);
            }
        }
        printf("%lld\n",b);
    }
    return 0;
}