# 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 secondtemplate<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;} `