Codeforces 241B Friends 字典树

Friends

首先确定第 m 大的是谁， 建出字典树之后二分去check， 找到第 m 大之后， 在跑一次字典树去统计总和。

为什么这个要取模啊， 卡了我半天。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5e4 + 10;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

const int LOG = 30;

int n, a[N];
LL m;
LL sum, num;

struct Trie {
    int ch[N * LOG][2], cnt[N * LOG];
    int f[N * LOG][30];
    int tot;
    void init() {
        tot = 1;
    }
    void ins(int x) {
        int u = 1;
        for(int i = LOG - 1; i >= 0; i--) {
            cnt[u]++;
            for(int j = LOG - 1; j >= 0; j--) f[u][j] += (x >> j & 1);
            if(!ch[u][x >> i & 1]) ch[u][x >> i & 1] = ++tot;
            u = ch[u][x >> i & 1];
        }
        for(int j = LOG - 1; j >= 0; j--) f[u][j] += (x >> j & 1);
        cnt[u]++;
    }
    PLL query(int x, int k, int op) {
        int u = 1;
        int now = 0;
        LL ans = 0;
        LL sum = 0;
        for(int i = LOG - 1; i >= 0 && u; i--) {
            if(x >> i & 1) {
                if(now + (1 << i) >= k) {
                    ans += cnt[ch[u][0]];
                    if(op) {
                        for(int j = LOG - 1; j >= 0; j--) {
                            if(x >> j & 1) sum += 1LL * (1 << j) * (cnt[ch[u][0]] - f[ch[u][0]][j]);
                            else sum += 1LL * (1 << j) * f[ch[u][0]][j];
                        }
                    }
                    u = ch[u][1];
                } else {
                    u = ch[u][0];
                    now += 1 << i;
                }
            } else {
                if(now + (1 << i) >= k) {
                    ans += cnt[ch[u][1]];
                    if(op) {
                        for(int j = LOG - 1; j >= 0; j--) {
                            if(x >> j & 1) sum += 1LL * (1 << j) * (cnt[ch[u][1]] - f[ch[u][1]][j]);
                            else sum += 1LL * (1 << j) * f[ch[u][1]][j];
                        }
                    }
                    u = ch[u][0];
                } else {
                    u = ch[u][1];
                    now += 1 << i;
                }
            }
        }
        if(now == k) ans += cnt[u], sum += 1LL * cnt[u] * now;
        return mk(ans, sum);
    }
} trie;

bool check(int x, int op) {
    sum = 0; num = 0;
    for(int i = 1; i <= n; i++) {
        PLL tmp = trie.query(a[i], x, op);
        num += tmp.fi; sum += tmp.se;
    }
    num /= 2; sum /= 2;
    return num >= m;
}


int main() {
    trie.init();
    scanf("%d%lld", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for(int i = 1; i <= n; i++) trie.ins(a[i]);
    int low = 0, high = (1 << 30) - 1, up = 0;
    while(low <= high) {
        int mid = low + high >> 1;
        if(check(mid, 0)) up = mid, low = mid + 1;
        else high = mid - 1;
    }
    check(up, 1);
    sum -= 1LL * (num - m) * up;
    printf("%lld", sum % mod);
    return 0;
}

/*
*/