CodeForces - 1417E XOR Inverse

传送门：CodeForces - 1417E

思路分析

比较两个数的大小，实际上就是比较二进制位第一个不同的位置，字典树保存相同前缀的位置
对于每一个二进制位，前缀相同的情况下，计算出这一位选0和1的逆序对数，分治下去

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 3e5+10;

int tree[N*30][2],cnt;
ll dp[N][2];
vector<int>v[N*30];

void ins(int num,int pos) {
    int x = 0;
    for(int i=30; i>=0; i--) {
        int s = (num>>i)&1;
        if(!tree[x][s]) tree[x][s]=++cnt;
        x = tree[x][s];
        v[x].pb(pos);
    }
}

void solve(int x,int pos) {
    if(pos == -1 ) return ;
    int l = tree[x][0];
    int r = tree[x][1];
    if(sz(v[l]) && sz(v[r])) {
        int sum = 0,j = 0;
        int n1 = sz(v[l]);
        int n2 = sz(v[r]);
        for(int i=0; i<n1; i++) {
            while(j<n2 && v[r][j]<v[l][i]) j++;
            dp[pos][0]+=j;
        }
        
        swap(v[l],v[r]);
        sum = 0,j = 0;
        n1 = sz(v[l]);
        n2 = sz(v[r]);
        for(int i=0; i<n1; i++) {
            while(j<n2 && v[r][j]<v[l][i]) j++;
            dp[pos][1]+=j;
        }
    }
    if(l) solve(l,pos-1);
    if(r) solve(r,pos-1); 
}


signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n;
    cin>>n;
    for(int i=1;i<=n;i++) {
        int x;
        cin>>x;
        ins(x,i);
    }
    solve(0,30);
    ll sum = 0,ans = 0;
    for(int i=30;i>=0;i--){
        if(dp[i][0]<=dp[i][1]){
            sum+=dp[i][0];
        }else{
            sum+=dp[i][1];
            ans|=(1LL<<i);
        }
    }
    cout<<sum<<" "<<ans<<endl;




    return 0;
}
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