【SRM】649 t2

題意

一個數列\(A\)，數的範圍均在\([0, 2^N-1]\)內，求一個\(B\)，使得新生成的數列\(C\)中逆序對最多（\(C_i = A_i xor B\)），輸出最多的逆序對。（\(|A|<=10^5\)）

分析

這種題固然要逐位考慮..考慮到二進制和xor，咱們須要想到trie...

題解

將數列插入到一棵trie，咱們在每個層記錄一個信息，表示\(B\)在這一層取\(0\)或取\(1\)新增的逆序對數，而後統計答案便可。
而因爲是xor操做，因此很好統計，咱們能夠每插入一個數就統計一次。

// BEGIN CUT HERE

// END CUT HERE
#line 5 "XorSequence.cpp"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
const int N=150005;
struct node {
    node *c[2];
    int s;
    void init() {
        c[0]=c[1]=0;
        s=0;
    }
}Po[N*31], *iT=Po, *root=0;
ll c[31][2];
node *newnode() {
    iT->init();
    return iT++;
}
void add(int w, int dep=29, node *&x=root) {
    if(!x) {
        x=newnode();
    }
    ++x->s;
    if(dep<0) {
        return;
    }
    int f=(w>>dep)&1;
    if(f) {
        add(w, dep-1, x->c[1]);
        if(x->c[0]) {
            c[dep][0]+=x->c[0]->s;
        }
    }
    else {
        add(w, dep-1, x->c[0]);
        if(x->c[1]) {
            c[dep][1]+=x->c[1]->s;
        }
    }
}
class XorSequence {
public:
    long long getmax(int N, int sz, int A0, int A1, int P, int Q, int R) {
        iT=Po;
        root=0;
        memset(c, 0, sizeof c);
        add(A0);
        for(int i=1; i<sz; ++i) {
            add(A1);
            int t=A1;
            A1=(1ll*A0*P+1ll*A1*Q+R)%N;
            A0=t;
        }
        ll ans=0;
        for(int i=0; i<30; ++i) {
            ans+=max(c[i][0], c[i][1]);
        }
        return ans;
    }

// BEGIN CUT HERE
    public:
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) test_case_5(); }
    private:
    template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
    void verify_case(int Case, const long long &Expected, const long long &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
    void test_case_0() { int Arg0 = 4; int Arg1 = 6; int Arg2 = 3; int Arg3 = 2; int Arg4 = 0; int Arg5 = 1; int Arg6 = 3; long long Arg7 = 8LL; verify_case(0, Arg7, getmax(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5, Arg6)); }
    void test_case_1() { int Arg0 = 8; int Arg1 = 8; int Arg2 = 2; int Arg3 = 5; int Arg4 = 3; int Arg5 = 1; int Arg6 = 4; long long Arg7 = 13LL; verify_case(1, Arg7, getmax(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5, Arg6)); }
    void test_case_2() { int Arg0 = 8; int Arg1 = 7; int Arg2 = 3; int Arg3 = 0; int Arg4 = 1; int Arg5 = 2; int Arg6 = 4; long long Arg7 = 12LL; verify_case(2, Arg7, getmax(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5, Arg6)); }
    void test_case_3() { int Arg0 = 32; int Arg1 = 15; int Arg2 = 7; int Arg3 = 9; int Arg4 = 11; int Arg5 = 2; int Arg6 = 1; long long Arg7 = 60LL; verify_case(3, Arg7, getmax(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5, Arg6)); }
    void test_case_4() { int Arg0 = 131072; int Arg1 = 131072; int Arg2 = 7; int Arg3 = 7; int Arg4 = 1; int Arg5 = 0; int Arg6 = 0; long long Arg7 = 0LL; verify_case(4, Arg7, getmax(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5, Arg6)); }
    void test_case_5() { int Arg0 = 131072; int Arg1 = 131070; int Arg2 = 411; int Arg3 = 415; int Arg4 = 398; int Arg5 = 463; int Arg6 = 9191; long long Arg7 = 4302679760LL; verify_case(5, Arg7, getmax(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5, Arg6)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main() {
    XorSequence ___test;
    ___test.run_test(-1);
    return 0;
}
// END CUT HERE