HackerEarth Xor sequence problem solution

In this HackerEarth Xor sequence problem solution, we have given a sequence a1...an. You need to perform m queries on it.

In each query, two parameters x, y are given, and we let

l = min(((x + ans.t) mod n) + 1, ((y + ans.t) mod n) + 1)

r = max(((x + ans.t) mod n) + 1, ((y + ans.t) mod n) + 1)

In this statement, t is a given constant, which can be 0 or 1. And ans is the answer of last query, with initial value is 0.

You need to find a pair (i,j), satisfies l <= i <= j <= r, and maximize ai xor ai+1 xor ... xor aj. The answer to this query is ai xor ai+1 xor ... xor aj.

HackerEarth Xor sequence problem solution.

#include<bits/stdc++.h>

typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef double lf;
typedef long double llf;
typedef std::pair<int,int> pii;

#define xx first
#define yy second

template<typename T> inline T max(T a,T b){return a>b?a:b;}
template<typename T> inline T min(T a,T b){return a<b?a:b;}
template<typename T> inline T abs(T a){return a>0?a:-a;}
template<typename T> inline bool repr(T &a,T b){return a<b?a=b,1:0;}
template<typename T> inline bool repl(T &a,T b){return a>b?a=b,1:0;}
template<typename T> inline T gcd(T a,T b){T t;if(a<b){while(a){t=a;a=b%a;b=t;}return b;}else{while(b){t=b;b=a%b;a=t;}return a;}}
template<typename T> inline T sqr(T x){return x*x;}
#define mp(a,b) std::make_pair(a,b)
#define pb push_back
#define I inline
#define mset(a,b) memset(a,b,sizeof(a))
#define mcpy(a,b) memcpy(a,b,sizeof(a))

#define fo0(i,n) for(int i=0,i##end=n;i<i##end;i++)
#define fo1(i,n) for(int i=1,i##end=n;i<=i##end;i++)
#define fo(i,a,b) for(int i=a,i##end=b;i<=i##end;i++)
#define fd0(i,n) for(int i=(n)-1;~i;i--)
#define fd1(i,n) for(int i=n;i;i--)
#define fd(i,a,b) for(int i=a,i##end=b;i>=i##end;i--)
#define foe(i,x)for(__typeof((x).end())i=(x).begin();i!=(x).end();++i)

struct Cg{I char operator()(){return getchar();}};
struct Cp{I void operator()(char x){putchar(x);}};
#define OP operator
#define RT return *this;
#define RX x=0;char t=P();while((t<'0'||t>'9')&&t!='-')t=P();bool f=0;\
if(t=='-')t=P(),f=1;x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0'
#define RL if(t=='.'){lf u=0.1;for(t=P();t>='0'&&t<='9';t=P(),u*=0.1)x+=u*(t-'0');}if(f)x=-x
#define RU x=0;char t=P();while(t<'0'||t>'9')t=P();x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0'
#define TR *this,x;return x;
I bool IS(char x){return x==10||x==13||x==' ';}template<typename T>struct Fr{T P;I Fr&OP,(int&x)
{RX;if(f)x=-x;RT}I OP int(){int x;TR}I Fr&OP,(ll &x){RX;if(f)x=-x;RT}I OP ll(){ll x;TR}I Fr&OP,(char&x)
{for(x=P();IS(x);x=P());RT}I OP char(){char x;TR}I Fr&OP,(char*x){char t=P();for(;IS(t);t=P());if(~t){for(;!IS
(t)&&~t;t=P())*x++=t;}*x++=0;RT}I Fr&OP,(lf&x){RX;RL;RT}I OP lf(){lf x;TR}I Fr&OP,(llf&x){RX;RL;RT}I OP llf()
{llf x;TR}I Fr&OP,(uint&x){RU;RT}I OP uint(){uint x;TR}I Fr&OP,(ull&x){RU;RT}I OP ull(){ull x;TR}};Fr<Cg>in;
#define WI(S) if(x){if(x<0)P('-'),x=-x;char s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
#define WL if(y){lf t=0.5;for(int i=y;i--;)t*=0.1;if(x>=0)x+=t;else x-=t,P('-');*this,(ll)(abs(x));P('.');if(x<0)\
x=-x;while(y--){x*=10;x-=floor(x*0.1)*10;P(((int)x)%10+'0');}}else if(x>=0)*this,(ll)(x+0.5);else *this,(ll)(x-0.5);
#define WU(S) if(x){char s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
template<typename T>struct Fw{T P;I Fw&OP,(int x){WI(10);RT}I Fw&OP()(int x){WI(10);RT}I Fw&OP,(uint x){WU(10);RT}
I Fw&OP()(uint x){WU(10);RT}I Fw&OP,(ll x){WI(19);RT}I Fw&OP()(ll x){WI(19);RT}I Fw&OP,(ull x){WU(20);RT}I Fw&OP()
(ull x){WU(20);RT}I Fw&OP,(char x){P(x);RT}I Fw&OP()(char x){P(x);RT}I Fw&OP,(const char*x){while(*x)P(*x++);RT}
I Fw&OP()(const char*x){while(*x)P(*x++);RT}I Fw&OP()(lf x,int y){WL;RT}I Fw&OP()(llf x,int y){WL;RT}};Fw<Cp>out;

const int N=20007,B=120,BM=155,BC=255,V=1<<29;

struct node
{
    int c[2],v;
}p[N*35];

int pm;

inline void modify(int&x,int t,int p)
{
    ::p[++pm]=::p[x],x=pm;
    ::p[x].v++;
    if(t)modify(::p[x].c[bool(p&t)],t>>1,p);
}

inline bool chk(int x,int y)
{
    return p[x].v-p[y].v;
}

inline int query(int x,int y,int u)
{
    int t=V,ans=0;
    for(;t;t>>=1)
        chk(p[x].c[!(u&t)],p[y].c[!(u&t)])?ans|=t,x=p[x].c[!(u&t)],y=p[y].c[!(u&t)]:(x=p[x].c[bool(u&t)],y=p[y].c[bool(u&t)]);
    return ans;
}

int n,m,type,s[N],r[N],f[BC][N];

int main()
{
    in,n,m,type;
    fo1(i,n)in,s[i];
    fo1(i,n)s[i]^=s[i-1];
    fd1(i,n+1)s[i]=s[i-1];
    ++n;
    fo1(i,n)r[i]=r[i-1],modify(r[i],V,s[i]);
    fo1(i,n/B)
    {
        int ans=0;
        fo(j,i*B,n)
        {
            repr(ans,query(r[j-1],r[i*B-1],s[j]));
            f[i][j]=ans;
        }
    }
    int l,r,ans=0;
    while(m--)
    {
        in,l,r;
        l=(l+type*ans)%(n-1)+1,r=(r+type*ans)%(n-1)+1;
        if(l>r)std::swap(l,r);++r;
        int lb=(l-1)/B+1,rb=(r-1)/B+1;
        if(lb==rb)
        {
            ans=0;
            fo(j,l,r)repr(ans,query(::r[j-1],::r[l-1],s[j]));
            out,ans,'\n';
        }
        else
        {
            ans=f[lb][r];
            fd(j,lb*B-1,l)repr(ans,query(::r[r],::r[j],s[j]));
            out,ans,'\n';
        }
    }
}

Second solution

#include <bits/stdc++.h>
using namespace std;

#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); --i)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
const int MOD = (int) 1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ll isqrt(ll k) {ll r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
#define db(x) cerr << #x << " = " << (x) << " ";
#define endln cerr << "\n";

const int maxn = 2e4 + 5;
const int logn = 31;
const int magic = 150;
int n, m, t;
int a[maxn];
int b[maxn];

int ptr;
int nxt[maxn * logn][2];

void clear() {
    FOR(i, 0, ptr + 1) {
        nxt[i][0] = nxt[i][1] = 0;
    }
    ptr = 0;
}

void add(int val) {
    int k = 0;
    FORd(i, logn, 0) {
        int x = bit(val, i);
        if (!nxt[k][x]) nxt[k][x] = ++ptr;
        k = nxt[k][x];
    }
}

int query(int val) {
    int k = 0, res = 0;
    FORd(i, logn, 0) {
        int x = bit(val, i);
        res <<= 1;
        if (nxt[k][x ^ 1]) {
            k = nxt[k][x ^ 1];
            res |= 1;
        }
        else {
            k = nxt[k][x];
        }
    }
    return res;
}

void chemthan() {
    cin >> n >> m >> t;
    assert(1 <= n && n <= 2e4);
    assert(1 <= m && m <= 2e4);
    assert(0 <= t && t <= 1);
    FOR(i, 1, n + 1) {
        cin >> a[i];
        assert(0 <= a[i] && a[i] <= 1e9);
        b[i] = a[i], a[i] ^= a[i - 1];
    }
    FORd(i, n + 1, 1) b[i] ^= b[i + 1];
    static int f[magic][maxn];
    static int g[magic][maxn];
    for (int st = 1; st <= n; st += magic) {
        clear();
        int ix = (st - 1) / magic;
        int mx = 0;
        FOR(i, st, n + 1) {
            add(a[i - 1]);
            chkmax(mx, query(a[i]));
            f[ix][i] = mx;
        }
    }
    for (int st = magic; st <= n; st += magic) {
        clear();
        int ix = st / magic - 1;
        int mx = 0;
        FORd(i, st + 1, 1) {
            add(b[i + 1]);
            chkmax(mx, query(b[i]));
            g[ix][i] = mx;
        }
    }
    int ans = 0;
    while (m--) {
        int x, y; cin >> x >> y;
        assert(0 <= x && x <= 1e9);
        assert(0 <= y && y <= 1e9);
        int l = (x + (long long) ans * t) % n + 1;
        int r = (y + (long long) ans * t) % n + 1;
        if (l > r) swap(l, r);
        if (r - l + 1 <= magic) {
            ans = 0;
            clear();
            FOR(i, l, r + 1) {
                add(a[i - 1]);
                chkmax(ans, query(a[i]));
            }
            cout << ans << "\n";
            continue;
        }
        int il = (l - 1 + magic - 1) / magic;
        int ir = r / magic - 1;
        ans = max(f[il][r], g[ir][l]);
        clear();
        FOR(i, l, il * magic + 1) {
            add(a[i - 1]);
        }
        FOR(i, (ir + 1) * magic + 1, r + 1) {
            chkmax(ans, query(a[i]));
        }
        cout << ans << "\n";
    }
}

int main(int argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
        assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
        assert(freopen(argv[2], "wb", stdout));
    }
    chemthan();
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}